Logarithms, Surd and Indices: CAT Previous Year Questions

Ans: A

The given inequality is:
$\left\{(a_{1})^{1}\times(a_{2})^{2}\times…\times(a_{20})^{20}\right\}<\left\{a_{21}\times a_{22}\times...\times a_{20+m}\right\}$

The left side of the inequality is:
$\left\{(a_{1})^{1}\times(a_{2})^{2}\times…\times(a_{20})^{20}\right\} = \left\{3^1\times3^4\times3^9…\times3^{400}\right\}$
This expression can be rewritten as:
$3^{(1^2 + 2^2 + \dots + 20^2)}$
The sum of the squares of the first $n$ natural numbers is given by the formula:
$\frac{n\cdot\left(n+1\right)\cdot\left(2n+1\right)}{6}$
For $n=20$, the sum is:
$\frac{20\cdot\left(20+1\right)\cdot\left(2\cdot20+1\right)}{6} = \frac{20\cdot21\cdot41}{6} = 2870$
So, the left side of the inequality simplifies to $3^{2870}$.

The right side of the inequality is:
$\left\{a_{21}\times a_{22}\times…\times a_{20+m}\right\} = 3^{21}\times3^{22}\times…\times3^{20+m}$
This can be written as:
$3^{(21+22+…+20+m)}$
We need to find the sum of the arithmetic progression $21+22+…+ (20+m)$.
Using the formula for the sum of the first $n$ natural numbers, $1+2+\cdots+n = \frac{n(n+1)}{2}$, we can express this sum as:
$[1+2+\cdots+(20+m)] – [1+2+\cdots+20]$
The sum of the first $(20+m)$ natural numbers is:
$\frac{(20+m)((20+m)+1)}{2} = \frac{(20+m)(21+m)}{2}$
The sum of the first 20 natural numbers is:
$\frac{20\cdot21}{2} = 210$
Therefore, the sum $21+22+\cdots+(20+m)$ is:
$\frac{(20+m)(21+m)}{2} – 210$
Expanding $(20+m)(21+m)$:
$(20+m)(21+m) = 420 + 20m + 21m + m^2 = m^2 + 41m + 420$
Substituting this back into the sum:
$\frac{m^2+41m+420}{2} – 210 = \frac{m^2+41m+420 – 420}{2} = \frac{m^2+41m}{2}$
So, the right side of the inequality is $3^{\frac{m^2+41m}{2}}$.

Now, we compare the powers since the bases are equal:
$2870 < \frac{m^2+41m}{2}$
Multiplying by 2:
$5740 < m^2+41m$
$5740 < m(m+41)$
We can test the given options to find the minimum value of $m$ that satisfies this inequality.

For $m=56$:
$56(56+41) = 56 \times 97 = 5432$.
$5740 < 5432$ is false.

For $m=57$:
$57(57+41) = 57 \times 98 = 5586$.
$5740 < 5586$ is false.

For $m=58$:
$58(58+41) = 58 \times 99 = 5742$.
$5740 < 5742$ is true.

Therefore, the minimum possible value for $m$ is 58.

Ans: D

For a logarithm with a base between 0 and 1 (specifically, 1/4), the expression $\log_{1/4}(x)$ will be positive only when $x$ is also between 0 and 1.

Given that $x$ is defined as $n^2 – 7n + 11$, and $n$ is an integer, the value of $x$ will always be an integer. An integer cannot fall strictly between 0 and 1.

Therefore, there is no integer value of $n$ that will satisfy the given inequality.

Ans: A

Given that $64 = 8^2$ and $512 = 8^3$. The equation is:
$\log_{64}{x^{2}+\log_{8}{\sqrt{y}+3\log_{512}{(\sqrt{y}z)}}}=4$

We utilize the logarithmic property: $\log_{b^m} a^n = \frac{n}{m} \log_b a$.
Substituting this into the equation:
$\log_{8^2}{x^{2}+\log_{8}{\sqrt{y}+3\log_{8^3}{(\sqrt{y}z)}}}=4$

Applying the property to the terms:
$\log_{8}{x}+\log_{8}{\sqrt{y}+\log_{8}{(\sqrt{y}z)}}=4$

Combining the logarithmic terms using the property $\log_b A + \log_b B = \log_b (AB)$:
$\log_8(x \cdot \sqrt{y} \cdot (\sqrt{y}z)) = 4$
$\log_8(xyz) = 4$

Converting the logarithmic equation to exponential form:
$xyz = 8^4$
$xyz = (2^3)^4$
$xyz = 2^{12}$

Now, employing the Arithmetic Mean – Geometric Mean (AM-GM) inequality, which states that for non-negative numbers, the arithmetic mean is greater than or equal to the geometric mean:
$\frac{x+y+z}{3} \ge \sqrt[3]{xyz}$

Substituting the value of $xyz$:
$\frac{x+y+z}{3} \ge \sqrt[3]{2^{12}}$
$\frac{x+y+z}{3} \ge 2^{12/3}$
$\frac{x+y+z}{3} \ge 2^4$
$\frac{x+y+z}{3} \ge 16$

Multiplying both sides by 3:
$x+y+z \ge 48$

Ans: D

For a logarithm to be defined, its base must be positive and not equal to 1. Therefore, $x$ must be greater than 3 and cannot be 4. Additionally, the argument of a logarithm must be positive, so $x^2 – 9 > 0$, which implies $x > 3$.

The given equation can be expressed as:
$\log_{x-3}{(x^{2}-9)}-\log_{x-3}{(x+1)} = 2$

Using the logarithm property $\log_b M – \log_b N = \log_b \frac{M}{N}$:
$\log_{x-3}{\dfrac{x^2-9}{x+1}} = 2$

Converting the logarithmic equation to an exponential form:
$\dfrac{x^2-9}{x+1} = (x-3)^2$

Factor the numerator:
$\dfrac{(x+3)(x-3)}{x+1} = (x-3)^2$

Assuming $x \neq 3$ (which is already covered by $x>3$), we can divide both sides by $(x-3)$:
$\dfrac{x+3}{x+1} = x-3$

Cross-multiply:
$x+3 = (x-3)(x+1)$

Expand the right side:
$x+3 = x^2 – 3x + x – 3$
$x+3 = x^2 – 2x – 3$

Rearrange into a quadratic equation:
$x^2 – 2x – x – 3 – 3 = 0$
$x^2 – 3x – 6 = 0$

The roots of this quadratic equation can be found using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$:
$x = \dfrac{-(-3) \pm \sqrt{(-3)^2 – 4(1)(-6)}}{2(1)}$
$x = \dfrac{3 \pm \sqrt{9 + 24}}{2}$
$x = \dfrac{3 \pm \sqrt{33}}{2}$

Considering the constraints established earlier ($x > 3$), the negative root is not valid. The positive root is $\dfrac{3+ \sqrt{33}}{2}$. This value is greater than 3, and since $\sqrt{33}$ is between 5 and 6, the value is approximately $(3+5.something)/2$, which is greater than 4. Thus, it satisfies the conditions.

The correct answer is option D.

Ans: 112

Explanation:We are given the equation:
$12^{12x}\times 4^{24x+12}\times 5^{2y}=8^{4z}\times 20 ^{12x} \times 243^{3x-6}$
First, we express all the bases in terms of their prime factors:
$12 = 2^2 \times 3$
$4 = 2^2$
$8 = 2^3$
$20 = 2^2 \times 5$
$243 = 3^5$

Substitute these into the equation:
$(2^2 \times 3)^{12x} \times (2^2)^{24x+12} \times 5^{2y} = (2^3)^{4z} \times (2^2 \times 5)^{12x} \times (3^5)^{3x-6}$

Now, apply the exponent rules $(a^m)^n = a^{mn}$ and $(ab)^m = a^m b^m$:
$(2^{2 \times 12x} \times 3^{12x}) \times 2^{2 \times (24x+12)} \times 5^{2y} = 2^{3 \times 4z} \times (2^{2 \times 12x} \times 5^{12x}) \times 3^{5 \times (3x-6)}$
$2^{24x} \times 3^{12x} \times 2^{48x+24} \times 5^{2y} = 2^{12z} \times 2^{24x} \times 5^{12x} \times 3^{15x-30}$

Combine the terms with the same base by adding their exponents:
$2^{24x + 48x + 24} \times 3^{12x} \times 5^{2y} = 2^{12z + 24x} \times 3^{15x-30} \times 5^{12x}$
$2^{72x + 24} \times 3^{12x} \times 5^{2y} = 2^{12z + 24x} \times 3^{15x-30} \times 5^{12x}$

For the equality to hold, the exponents of each prime base on both sides of the equation must be equal.

Equating the exponents of base 2:
$72x + 24 = 12z + 24x$
$72x – 24x + 24 = 12z$
$48x + 24 = 12z$
Divide by 12:
$4x + 2 = z$ (Equation 1)

Equating the exponents of base 3:
$12x = 15x – 30$
$30 = 15x – 12x$
$30 = 3x$
$x = 10$

Equating the exponents of base 5:
$2y = 12x$
Divide by 2:
$y = 6x$

Now substitute the value of x into Equation 1 to find z:
$z = 4(10) + 2$
$z = 40 + 2$
$z = 42$

Now substitute the value of x into the equation for y:
$y = 6(10)$
$y = 60$

We are given that x, y, and z are natural numbers. Our calculated values are x=10, y=60, and z=42, which are all natural numbers.

We need to find $x + y + z$:
$x + y + z = 10 + 60 + 42$
$x + y + z = 112$

Final_Answer:112

Ans: C

Explanation:The given equation is:
$\left(\cfrac{1}{8}\right)^{k}\times \left(\cfrac{1}{32768}\right)^{\cfrac{1}{3}}=\cfrac{1}{8}\times \left(\cfrac{1}{32768}\right)^{\cfrac{1}{k}}$

We can rewrite the terms using powers of 2.
$8 = 2^3$
$32768 = 2^{15}$

So, the equation becomes:
$\left(2^{-3}\right)^{k}\times \left(2^{-15}\right)^{\cfrac{1}{3}}=2^{-3}\times \left(2^{-15}\right)^{\cfrac{1}{k}}$
$2^{-3k}\times 2^{-\frac{15}{3}}=2^{-3}\times 2^{-\frac{15}{k}}$
$2^{-3k}\times 2^{-5}=2^{-3}\times 2^{-\frac{15}{k}}$

Now, combine the terms on each side using the property $a^m \times a^n = a^{m+n}$:
$2^{-3k-5}=2^{-3-\frac{15}{k}}$

Since the bases are equal, the exponents must be equal:
$-3k-5 = -3-\frac{15}{k}$

We need to solve this equation for k. First, move all terms to one side:
$-3k-5+3+\frac{15}{k} = 0$
$-3k-2+\frac{15}{k} = 0$

Multiply the entire equation by k to eliminate the fraction. Note that $k \neq 0$ because it appears in the denominator.
$-3k^2-2k+15 = 0$

Multiply by -1 to make the leading coefficient positive:
$3k^2+2k-15 = 0$

This is a quadratic equation. We can solve for k using the quadratic formula $k = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, where $a=3$, $b=2$, and $c=-15$.
$k = \frac{-2 \pm \sqrt{2^2 – 4(3)(-15)}}{2(3)}$
$k = \frac{-2 \pm \sqrt{4 + 180}}{6}$
$k = \frac{-2 \pm \sqrt{184}}{6}$
$k = \frac{-2 \pm \sqrt{4 \times 46}}{6}$
$k = \frac{-2 \pm 2\sqrt{46}}{6}$
$k = \frac{-1 \pm \sqrt{46}}{3}$

The real values of k are $k_1 = \frac{-1 + \sqrt{46}}{3}$ and $k_2 = \frac{-1 – \sqrt{46}}{3}$.
The question asks for the sum of all real values of k.
Sum of roots $= k_1 + k_2 = \frac{-1 + \sqrt{46}}{3} + \frac{-1 – \sqrt{46}}{3} = \frac{-1 + \sqrt{46} – 1 – \sqrt{46}}{3} = \frac{-2}{3}$.

Alternatively, for a quadratic equation $ax^2+bx+c=0$, the sum of the roots is given by $-\frac{b}{a}$.
In our equation $3k^2+2k-15=0$, $a=3$ and $b=2$.
Sum of the real values of k $= -\frac{2}{3}$.

Let’s double check the calculations.
$8 = 2^3$
$32768 = 2^{15}$
Equation: $(2^{-3})^k \times (2^{-15})^{1/3} = 2^{-3} \times (2^{-15})^{1/k}$
$2^{-3k} \times 2^{-5} = 2^{-3} \times 2^{-15/k}$
$2^{-3k-5} = 2^{-3 – 15/k}$
$-3k-5 = -3 – 15/k$
$-3k-2 = -15/k$
$-3k^2 – 2k = -15$
$3k^2 + 2k – 15 = 0$

The sum of the roots of $3k^2+2k-15=0$ is $-\frac{b}{a} = -\frac{2}{3}$.
The problem statement mentions “real values of k”. We should verify if the roots are indeed real. The discriminant of the quadratic equation $3k^2+2k-15=0$ is $\Delta = b^2-4ac = (2)^2 – 4(3)(-15) = 4 + 180 = 184$. Since $\Delta > 0$, the roots are real and distinct.

Thus, the sum of all real values of k is $-\frac{2}{3}$.

Correct_Option: C

Ans: 31

By applying the rule of logarithms where $\log_{a^p}b = \frac{1}{p} \log_a b$, the expression
$4 \log_{10} x + 4 \log_{100} x + 8 \log_{1000} x = 13$
transforms into
$4\log_{10}x+2\log_{10}x+\frac{8}{3}\log_{10}x=13$
This simplifies to
$\frac{26}{3}\log_{10}x=13$
Solving for the logarithm gives
$\log_{10}x=1.5$
Therefore, the value of x is
$x=10^{1.5}$
which is equivalent to
$x=\sqrt{1000}$
Applying the Greatest Integer Function, denoted by [.] as specified in the problem, we find
$\left[\sqrt{1000}\right]=31$
This result, 31, represents the largest integer less than or equal to x.

Ans: A

The expression $(a + b \sqrt{n})$ represents the positive square root of $(29 – 12\sqrt{5})$.
Therefore, we can write:
$(29 – 12\sqrt{5}) = (a + b \sqrt{n})^2$
Expanding the right side:
$29 – 12\sqrt{5} = a^2 + b^2n + 2ab\sqrt{n}$
By comparing the rational and irrational parts, we get two equations:
$a^2 + b^2n = 29$ (Equation 1)
$2ab\sqrt{n} = -12\sqrt{5}$ (Equation 2)

From Equation 2, we can simplify:
$ab\sqrt{n} = -6\sqrt{5}$
Squaring both sides of this simplified equation:
$(ab\sqrt{n})^2 = (-6\sqrt{5})^2$
$a^2b^2n = 36 \times 5$
$a^2b^2n = 180$

From this, we can express $b^2n$ in terms of $a^2$:
$b^2n = \frac{180}{a^2}$

Substitute this expression for $b^2n$ into Equation 1:
$a^2 + \frac{180}{a^2} = 29$
Multiply the entire equation by $a^2$ to eliminate the fraction:
$a^4 + 180 = 29a^2$
Rearrange into a quadratic form in terms of $a^2$:
$a^4 – 29a^2 + 180 = 0$

Let $x = a^2$. The equation becomes a quadratic equation:
$x^2 – 29x + 180 = 0$
We can solve this quadratic equation for $x$ (which is $a^2$) using the quadratic formula:
$x = \frac{-(-29) \pm \sqrt{(-29)^2 – 4(1)(180)}}{2(1)}$
$a^2 = \frac{29 \pm \sqrt{841 – 720}}{2}$
$a^2 = \frac{29 \pm \sqrt{121}}{2}$
$a^2 = \frac{29 \pm 11}{2}$
This gives two possible values for $a^2$:
$a^2 = \frac{29 + 11}{2} = \frac{40}{2} = 20$
or
$a^2 = \frac{29 – 11}{2} = \frac{18}{2} = 9$

So, the possible values for $a^2$ are 9 and 20. Since $a^2 + b^2n = 29$, if $a^2=9$, then $b^2n=20$, and if $a^2=20$, then $b^2n=9$. This means that the pair $(a^2, b^2n)$ is either $(9, 20)$ or $(20, 9)$.

From the condition $ab\sqrt{n} = -6\sqrt{5}$, we know that the product $ab$ must be negative. This implies that one of $a$ or $b$ is positive, and the other is negative.
We are looking for the positive square root, $(a + b\sqrt{n})$.
We also need to consider the goal of maximizing the values of $a$, $b$, and $n$.

Let’s examine the possible pairs for $(a^2, b^2n)$:

Case 1: $a^2 = 9$ and $b^2n = 20$
From $a^2=9$, $a$ can be 3 or -3.
From $b^2n=20$, we need to find integer values for $b$ and $n$ such that $b^2n=20$.
If we consider $a=-3$ (to make $ab$ negative), then from $ab\sqrt{n} = -6\sqrt{5}$, we have $(-3)b\sqrt{n} = -6\sqrt{5}$, so $b\sqrt{n} = 2\sqrt{5}$.
Squaring this, $b^2n = (2\sqrt{5})^2 = 4 \times 5 = 20$. This matches our condition for $b^2n$.
To find $b$ and $n$, we have $b\sqrt{n} = 2\sqrt{5}$. If we choose $n=5$, then $b\sqrt{5} = 2\sqrt{5}$, which means $b=2$.
In this case, $a=-3$, $b=2$, and $n=5$.
The square root is $a+b\sqrt{n} = -3 + 2\sqrt{5}$. To verify, $(-3+2\sqrt{5})^2 = 9 + 8 – 12\sqrt{5} = 17 – 12\sqrt{5}$. This does not match $29 – 12\sqrt{5}$.

Let’s reconsider $b\sqrt{n} = 2\sqrt{5}$. We need to extract rational $b$ and a square-free $n$.
If we set $n=5$, then $b=2$. $a^2=9$, $b^2n = 2^2 \times 5 = 4 \times 5 = 20$. $a^2+b^2n = 9+20=29$. $ab\sqrt{n} = (-3)(2)\sqrt{5} = -6\sqrt{5}$. This works with $a=-3, b=2, n=5$. The value of $a+b\sqrt{n} = -3+2\sqrt{5}$. This is a valid square root.

Case 2: $a^2 = 20$ and $b^2n = 9$
From $a^2=20$, $a$ can be $\sqrt{20} = 2\sqrt{5}$ or $-\sqrt{20} = -2\sqrt{5}$.
From $b^2n=9$, $b$ can be 3 or -3, and $n$ must be a perfect square or 1 for $b$ to be rational and $n$ to be the radicand.
Let’s try to get an integer value for $a$ by choosing $n$ appropriately.

Let’s go back to $ab\sqrt{n} = -6\sqrt{5}$. This implies that $\sqrt{n}$ must be related to $\sqrt{5}$.
Let’s assume $n=5$.
Then $ab\sqrt{5} = -6\sqrt{5}$, which means $ab = -6$.

Now we combine this with the pairs for $(a^2, b^2n)$:

Scenario A: $a^2=9$ and $b^2n=20$.
If $n=5$, then $b^2 \times 5 = 20 \implies b^2 = 4 \implies b = \pm 2$.
If $a^2=9$, then $a = \pm 3$.
We need $ab=-6$.
If $a=3$, then $3b=-6 \implies b=-2$. So, $a=3, b=-2, n=5$.
The square root is $a+b\sqrt{n} = 3 + (-2)\sqrt{5} = 3 – 2\sqrt{5}$.
Let’s check: $(3-2\sqrt{5})^2 = 9 + 8 – 12\sqrt{5} = 17 – 12\sqrt{5}$. This is incorrect.

If $a=-3$, then $(-3)b=-6 \implies b=2$. So, $a=-3, b=2, n=5$.
The square root is $a+b\sqrt{n} = -3 + 2\sqrt{5}$.
Let’s check: $(-3+2\sqrt{5})^2 = 9 + 8 – 12\sqrt{5} = 17 – 12\sqrt{5}$. This is incorrect.

There seems to be a misunderstanding in how $a$ and $b$ are treated when $n$ is not a perfect square.

Let’s re-evaluate $(a + b \sqrt{n})^2 = a^2 + b^2n + 2ab\sqrt{n}$.
We have $29 – 12\sqrt{5}$.
The structure suggests that $n=5$.
So, $(a + b\sqrt{5})^2 = a^2 + 5b^2 + 2ab\sqrt{5}$.
Comparing this to $29 – 12\sqrt{5}$, we get:
$a^2 + 5b^2 = 29$ (Equation 3)
$2ab = -12 \implies ab = -6$ (Equation 4)

From Equation 4, $b = -6/a$. Substitute this into Equation 3:
$a^2 + 5(-6/a)^2 = 29$
$a^2 + 5(36/a^2) = 29$
$a^2 + 180/a^2 = 29$
Multiply by $a^2$:
$a^4 + 180 = 29a^2$
$a^4 – 29a^2 + 180 = 0$

This leads to the same quadratic for $a^2$:
$a^2 = 9 \text{ or } 20$

If $a^2 = 9$, then $a = \pm 3$.
If $a=3$, then from $ab=-6$, $3b=-6 \implies b=-2$.
The expression is $a+b\sqrt{n} = 3 + (-2)\sqrt{5} = 3 – 2\sqrt{5}$.
$(3 – 2\sqrt{5})^2 = 9 + 4(5) – 12\sqrt{5} = 9 + 20 – 12\sqrt{5} = 29 – 12\sqrt{5}$. This matches!
Here, $a=3$, $b=-2$, and $n=5$.

If $a=-3$, then from $ab=-6$, $(-3)b=-6 \implies b=2$.
The expression is $a+b\sqrt{n} = -3 + 2\sqrt{5}$.
$(-3 + 2\sqrt{5})^2 = 9 + 4(5) – 12\sqrt{5} = 9 + 20 – 12\sqrt{5} = 29 – 12\sqrt{5}$. This also matches!
Here, $a=-3$, $b=2$, and $n=5$.

If $a^2 = 20$, then $a = \pm \sqrt{20} = \pm 2\sqrt{5}$.
If $a = 2\sqrt{5}$, then from $ab=-6$, $(2\sqrt{5})b=-6 \implies b = -6/(2\sqrt{5}) = -3/\sqrt{5} = -3\sqrt{5}/5$.
Here, $a = 2\sqrt{5}$ and $b = -3\sqrt{5}/5$.
The expression is $a+b\sqrt{n} = 2\sqrt{5} + (-3\sqrt{5}/5)\sqrt{5} = 2\sqrt{5} – 15/5 = 2\sqrt{5} – 3$.
$(2\sqrt{5} – 3)^2 = (2\sqrt{5})^2 + 3^2 – 2(2\sqrt{5})(3) = 20 + 9 – 12\sqrt{5} = 29 – 12\sqrt{5}$. This also matches!
Here, $a=2\sqrt{5}$, $b=-3\sqrt{5}/5$, and $n=5$. This is not in the form $a+b\sqrt{n}$ where $a, b, n$ are integers (as typically implied unless stated otherwise).

If $a = -2\sqrt{5}$, then from $ab=-6$, $(-2\sqrt{5})b=-6 \implies b = -6/(-2\sqrt{5}) = 3/\sqrt{5} = 3\sqrt{5}/5$.
The expression is $a+b\sqrt{n} = -2\sqrt{5} + (3\sqrt{5}/5)\sqrt{5} = -2\sqrt{5} + 15/5 = -2\sqrt{5} + 3$.
$(-2\sqrt{5} + 3)^2 = (3 – 2\sqrt{5})^2 = 29 – 12\sqrt{5}$. This also matches.
Here, $a=-2\sqrt{5}$, $b=3\sqrt{5}/5$, and $n=5$.

The problem states $(a + b \sqrt{n})$ is the positive square root.
We found two square roots: $3 – 2\sqrt{5}$ and $-3 + 2\sqrt{5}$.
Since $\sqrt{5}$ is approximately 2.236:
$3 – 2\sqrt{5} \approx 3 – 2(2.236) = 3 – 4.472 = -1.472$ (negative)
$-3 + 2\sqrt{5} \approx -3 + 2(2.236) = -3 + 4.472 = 1.472$ (positive)
So, the positive square root is $-3 + 2\sqrt{5}$.
This means $a = -3$, $b = 2$, and $n = 5$.

Let’s consider the values: $a=-3$, $b=2$, $n=5$.
We need to maximize $a, b, n$. However, these values are fixed once the expression is identified. The phrasing “maximise the value of a, b and n” might be misleading or imply a choice if multiple integer solutions exist.

Let’s assume the problem implies that $a, b, n$ are integers. In that case, the only integer solution for $a$ and $b$ when $n=5$ are from $a^2=9$:
Possibility 1: $a=3, b=-2, n=5$. Square root: $3 – 2\sqrt{5}$ (negative).
Possibility 2: $a=-3, b=2, n=5$. Square root: $-3 + 2\sqrt{5}$ (positive).

The positive square root is $-3 + 2\sqrt{5}$.
So, $a = -3$, $b = 2$, $n = 5$.
The sum $a+b+n = -3 + 2 + 5 = 4$.

Let’s re-read the original explanation:
“We can have a=-3, b=1 and n=20.
This satisfies all the above equations, and the value of a+b+n=18.”

This implies a different interpretation of $a, b, n$.
Let’s go back to $a^2=9 \text{ or } 20$ and $b^2n=20 \text{ or } 9$.
And $ab\sqrt{n}=-6\sqrt{5}$.

If $n=20$, then $ab\sqrt{20} = -6\sqrt{5}$
$ab \cdot 2\sqrt{5} = -6\sqrt{5}$
$2ab = -6 \implies ab = -3$.

Now, let’s check the pairs for $(a^2, b^2n)$:

Case 1: $a^2=9$ and $b^2n=20$.
If $n=20$, then $b^2(20) = 20 \implies b^2 = 1 \implies b = \pm 1$.
If $a^2=9$, then $a = \pm 3$.
We need $ab=-3$.
If $a=3$, then $3b=-3 \implies b=-1$. So, $a=3, b=-1, n=20$.
Square root: $a+b\sqrt{n} = 3 + (-1)\sqrt{20} = 3 – \sqrt{20} = 3 – 2\sqrt{5}$.
Check: $(3 – 2\sqrt{5})^2 = 29 – 12\sqrt{5}$. This works.
The values are $a=3, b=-1, n=20$. The sum $a+b+n = 3 – 1 + 20 = 22$.

If $a=-3$, then $(-3)b=-3 \implies b=1$. So, $a=-3, b=1, n=20$.
Square root: $a+b\sqrt{n} = -3 + 1\sqrt{20} = -3 + \sqrt{20} = -3 + 2\sqrt{5}$.
Check: $(-3 + 2\sqrt{5})^2 = 29 – 12\sqrt{5}$. This works.
The values are $a=-3, b=1, n=20$. The sum $a+b+n = -3 + 1 + 20 = 18$.

Case 2: $a^2=20$ and $b^2n=9$.
If $n=20$, then $b^2(20) = 9 \implies b^2 = 9/20$. $b = \pm 3/\sqrt{20} = \pm 3/(2\sqrt{5})$. Not integer.

The explanation’s conclusion that “a=-3, b=1 and n=20” leads to “a+b+n=18” seems correct based on the derivation.
This choice of $a, b, n$ results in the positive square root $-3 + 1\sqrt{20} = -3 + 2\sqrt{5}$.

The phrasing “maximise the value of a, b and n” is still a bit ambiguous if integer values are assumed. If we want to maximize the sum $a+b+n$, then $a=3, b=-1, n=20$ gives sum 22, while $a=-3, b=1, n=20$ gives sum 18.

However, the original explanation explicitly states “We can have a=-3, b=1 and n=20. This satisfies all the above equations, and the value of a+b+n=18.” This implies that this specific combination leads to the desired outcome. The fact that it’s the positive square root is key.

$a=-3, b=1, n=20$ yields $-3 + 1\sqrt{20} = -3 + 2\sqrt{5}$, which is the positive root.
If we chose $a=3, b=-1, n=20$, it yields $3 – 1\sqrt{20} = 3 – 2\sqrt{5}$, which is the negative root.

Therefore, the selection of $a=-3, b=1, n=20$ is uniquely determined by the condition that it represents the positive square root. The maximization part might be a distraction or context from the original CAT question’s options.

The core logic is:
1. Set up the equality: $(29 – 12\sqrt{5}) = (a + b \sqrt{n})^2$.
2. Expand and equate rational and irrational parts.
3. Solve the system of equations for $a^2, b^2n$ and $ab\sqrt{n}$.
4. Identify possible integer values for $a, b, n$.
5. Select the combination that gives the positive square root.

$29 – 12\sqrt{5} = (a + b\sqrt{n})^2 = a^2 + b^2n + 2ab\sqrt{n}$
Comparing coefficients:
$a^2 + b^2n = 29$
$2ab\sqrt{n} = -12\sqrt{5} \implies ab\sqrt{n} = -6\sqrt{5}$

Squaring the second equation:
$(ab\sqrt{n})^2 = (-6\sqrt{5})^2$
$a^2b^2n = 36 \times 5 = 180$

Let’s consider possible integer values for $n$ that allow for integer $a$ and $b$ in $ab\sqrt{n} = -6\sqrt{5}$.
If $n=5$, then $ab\sqrt{5} = -6\sqrt{5} \implies ab = -6$.
And $a^2 + 5b^2 = 29$.
From $ab=-6$, $b=-6/a$.
$a^2 + 5(-6/a)^2 = 29$
$a^2 + 5(36/a^2) = 29$
$a^2 + 180/a^2 = 29$
$a^4 – 29a^2 + 180 = 0$
$a^2 = 9 \text{ or } 20$
If $a^2=9$, $a=\pm 3$.
If $a=3$, $b=-2$. Square root: $3 – 2\sqrt{5}$ (negative).
If $a=-3$, $b=2$. Square root: $-3 + 2\sqrt{5}$ (positive).
So, for $n=5$, the positive root implies $a=-3, b=2, n=5$. Sum $a+b+n = -3+2+5 = 4$.

If $n=20$, then $ab\sqrt{20} = -6\sqrt{5}$
$ab(2\sqrt{5}) = -6\sqrt{5}$
$2ab = -6 \implies ab = -3$
And $a^2 + b^2(20) = 29$
From $ab=-3$, $b=-3/a$.
$a^2 + 20(-3/a)^2 = 29$
$a^2 + 20(9/a^2) = 29$
$a^2 + 180/a^2 = 29$
$a^4 – 29a^2 + 180 = 0$
$a^2 = 9 \text{ or } 20$
If $a^2=9$, $a=\pm 3$.
If $a=3$, $b=-1$. Square root: $3 – 1\sqrt{20} = 3 – 2\sqrt{5}$ (negative).
If $a=-3$, $b=1$. Square root: $-3 + 1\sqrt{20} = -3 + 2\sqrt{5}$ (positive).
So, for $n=20$, the positive root implies $a=-3, b=1, n=20$. Sum $a+b+n = -3+1+20 = 18$.

The original explanation uses the case $n=20$ leading to the sum 18. This is the most likely intended interpretation.

Let’s rephrase the explanation following the structure.

The value $(a + b \sqrt{n})$ represents the positive square root of $(29 – 12\sqrt{5})$.
This implies:
$(29 – 12\sqrt{5}) = (a + b \sqrt{n})^2$
Expanding the right side:
$29 – 12\sqrt{5} = a^2 + b^2n + 2ab\sqrt{n}$
By equating the rational and irrational components on both sides, we obtain:
$a^2 + b^2n = 29$ (Equation 1)
$2ab\sqrt{n} = -12\sqrt{5}$ (Equation 2)

Simplifying Equation 2 gives:
$ab\sqrt{n} = -6\sqrt{5}$
Squaring both sides of this simplified equation yields:
$(ab\sqrt{n})^2 = (-6\sqrt{5})^2$
$a^2b^2n = 36 \times 5$
$a^2b^2n = 180$

From this relationship, we can express $b^2n$ in terms of $a^2$:
$b^2n = \frac{180}{a^2}$

Substitute this expression for $b^2n$ back into Equation 1:
$a^2 + \frac{180}{a^2} = 29$
To eliminate the fraction, multiply the entire equation by $a^2$:
$a^4 + 180 = 29a^2$
Rearranging the terms to form a quadratic equation in terms of $a^2$:
$a^4 – 29a^2 + 180 = 0$

Let $x = a^2$. The equation becomes a quadratic equation in $x$:
$x^2 – 29x + 180 = 0$
Solving this quadratic equation for $x$ using the quadratic formula:
$x = \frac{-(-29) \pm \sqrt{(-29)^2 – 4(1)(180)}}{2(1)}$
$a^2 = \frac{29 \pm \sqrt{841 – 720}}{2}$
$a^2 = \frac{29 \pm \sqrt{121}}{2}$
$a^2 = \frac{29 \pm 11}{2}$
This gives two possible values for $a^2$:
$a^2 = \frac{29 + 11}{2} = \frac{40}{2} = 20$
or
$a^2 = \frac{29 – 11}{2} = \frac{18}{2} = 9$

Thus, $a^2$ can be either 9 or 20. Consequently, the pair $(a^2, b^2n)$ must be either $(9, 20)$ or $(20, 9)$.

From the equation $ab\sqrt{n} = -6\sqrt{5}$, we know that the product $ab$ must be negative, meaning $a$ and $b$ have opposite signs.
We are looking for the positive square root $(a + b \sqrt{n})$.

Let’s consider integer values for $a, b, n$.
If we set $n=20$, then the condition $ab\sqrt{n} = -6\sqrt{5}$ becomes:
$ab\sqrt{20} = -6\sqrt{5}$
$ab(2\sqrt{5}) = -6\sqrt{5}$
Dividing both sides by $2\sqrt{5}$ gives:
$ab = -3$

Now we check our possible values for $a^2$:
If $a^2 = 9$, then $a = \pm 3$.
If $a = 3$, then from $ab = -3$, we get $3b = -3 \implies b = -1$.
This combination is $a=3, b=-1, n=20$. The square root is $3 + (-1)\sqrt{20} = 3 – \sqrt{20} = 3 – 2\sqrt{5}$.
Calculating the value: $3 – 2\sqrt{5} \approx 3 – 2(2.236) = 3 – 4.472 = -1.472$ (negative).

If $a = -3$, then from $ab = -3$, we get $(-3)b = -3 \implies b = 1$.
This combination is $a=-3, b=1, n=20$. The square root is $-3 + 1\sqrt{20} = -3 + \sqrt{20} = -3 + 2\sqrt{5}$.
Calculating the value: $-3 + 2\sqrt{5} \approx -3 + 2(2.236) = -3 + 4.472 = 1.472$ (positive).

This combination $a=-3, b=1, n=20$ satisfies all the derived conditions, including yielding the positive square root.
The sum of these values is $a + b + n = -3 + 1 + 20 = 18$.

If we consider $a^2 = 20$, then $a = \pm \sqrt{20} = \pm 2\sqrt{5}$. This would not give integer values for $a$ if $n=20$.

The combination $a=-3, b=1, n=20$ is selected because it fulfills the requirement of representing the positive square root. The sum of these integer values is 18.

Ans: 14

The initial segment of the equation can be transformed as follows:

Start with:
$ \frac{\frac{1}{3}\log_2\left(a+b\right)}{\log_2c} $

Apply the rule $ m \log_x y = \log_x y^m $ to the numerator:
$ \frac{\log_2\left(a+b\right)^{\frac{1}{3}}}{\log_2c} $

Utilize the change of base formula for logarithms, $\frac{\log_x y}{\log_x z} = \log_z y$:
$ \log_c\left(a+b\right)^{\frac{1}{3}} $

Applying the same sequence of operations to the second term yields:
$ \log_c\left(a-b\right)^{\frac{1}{3}} $

The left side of the equation thus becomes:
$ \log_c\left(a+b\right)^{\frac{1}{3}}+\log_c\left(a-b\right)^{\frac{1}{3}} $

Using the property $\log_x y + \log_x z = \log_x (yz)$:
$ \log_c\left[\left(a+b\right)^{\frac{1}{3}}\left(a-b\right)^{\frac{1}{3}}\right] $

Combine the terms within the exponent:
$ \log_c\left[\left(a+b\right)\left(a-b\right)\right]^{\frac{1}{3}} $

Simplify the expression inside the brackets:
$ \log_c\left[\left(a^2-b^2\right)\right]^{\frac{1}{3}} $

This expression is stated to be equal to $\frac{2}{3}$.

By the definition of logarithms, if $\log_N M = x$, then $N^x = M$. Applying this here:
$ c^{\frac{2}{3}}=\left(a^2-b^2\right)^{\frac{1}{3}} $

Cube both sides of the equation:
$ \left(c^{\frac{2}{3}}\right)^3 = \left[\left(a^2-b^2\right)^{\frac{1}{3}}\right]^3 $
$ c^2 = a^2-b^2 $

Rearrange to solve for $a^2$:
$ a^2 = b^2+c^2 $

We are given that $b \le 10$ and $c \le 10$, and we aim to maximize $a^2$.
This relationship resembles the Pythagorean theorem. To maximize $a^2$, we should maximize $b$ and $c$. Setting $b=10$ and $c=10$ gives $a^2 = 10^2 + 10^2 = 100 + 100 = 200$.

However, we need to find an integer value for $a$. We look for the largest perfect square less than 200. This is $196$, which is $14^2$.
Therefore, the maximum integer value for $a$ is 14.

Ans: B

Explanation:We are given the equation $(a + b\sqrt{3})^2 = 52 + 30\sqrt{3}$, where a and b are natural numbers.
Let’s expand the left side of the equation:
$(a + b\sqrt{3})^2 = a^2 + 2(a)(b\sqrt{3}) + (b\sqrt{3})^2$
$(a + b\sqrt{3})^2 = a^2 + 2ab\sqrt{3} + b^2(3)$
$(a + b\sqrt{3})^2 = a^2 + 3b^2 + 2ab\sqrt{3}$

Now, we equate this to the right side of the given equation:
$a^2 + 3b^2 + 2ab\sqrt{3} = 52 + 30\sqrt{3}$

Since a and b are natural numbers, we can equate the rational and irrational parts of the equation:
1. Rational part: $a^2 + 3b^2 = 52$
2. Irrational part: $2ab\sqrt{3} = 30\sqrt{3}$

From the irrational part, we can divide both sides by $\sqrt{3}$:
$2ab = 30$
Dividing by 2, we get:
$ab = 15$

Now we need to find natural numbers a and b that satisfy both $ab = 15$ and $a^2 + 3b^2 = 52$.
The pairs of natural numbers (a, b) such that $ab = 15$ are:
(1, 15), (3, 5), (5, 3), (15, 1).

Let’s test each pair in the equation $a^2 + 3b^2 = 52$:

Case 1: a = 1, b = 15
$a^2 + 3b^2 = 1^2 + 3(15^2) = 1 + 3(225) = 1 + 675 = 676 \neq 52$

Case 2: a = 3, b = 5
$a^2 + 3b^2 = 3^2 + 3(5^2) = 9 + 3(25) = 9 + 75 = 84 \neq 52$

Case 3: a = 5, b = 3
$a^2 + 3b^2 = 5^2 + 3(3^2) = 25 + 3(9) = 25 + 27 = 52$
This pair satisfies both equations.

Case 4: a = 15, b = 1
$a^2 + 3b^2 = 15^2 + 3(1^2) = 225 + 3(1) = 225 + 3 = 228 \neq 52$

So, the natural numbers are $a = 5$ and $b = 3$.
We are asked to find $a + b$.
$a + b = 5 + 3 = 8$

The value of $a+b$ is 8.

The options are:
Option A: 7
Option B: 8
Option C: 9
Option D: 10

Our calculated value matches Option B.

Correct_Option:B

Ans: A

Let’s represent the expression $10^x$ as $y$.
Substituting this into the given equation, we have:
$y + \frac{4}{y} = \frac{81}{2}$
To solve for $y$, we can rearrange this into a quadratic equation:
$2y^2 – 81y + 8 = 0$
We are looking for the sum of the possible values of $x$. Let these values be $x_1$ and $x_2$.
These values of $x$ correspond to the logarithms of the possible values of $y$. Specifically, if $y_1$ and $y_2$ are the roots of the quadratic equation, then $x_1 = \log_{10} y_1$ and $x_2 = \log_{10} y_2$.
The sum of the possible values of $x$ is $x_1 + x_2$, which can be written as $\log_{10} y_1 + \log_{10} y_2$.
Using the properties of logarithms, this sum is equal to $\log_{10} (y_1 \times y_2)$.
From the quadratic equation $2y^2 – 81y + 8 = 0$, the product of the roots ($y_1 \times y_2$) is given by the constant term divided by the coefficient of the $y^2$ term, which is $\frac{8}{2} = 4$.
Therefore, the sum of the values of $x$ is $\log_{10} 4$.
This can be further simplified as $2 \log_{10} 2$.
Thus, the correct option is A.

Ans: 2

Explanation:We are given the following equations:
1) $3^a = 4$
2) $4^b = 5$
3) $5^c = 6$
4) $6^d = 7$
5) $7^e = 8$
6) $8^f = 9$

We want to find the value of the product $abcdef$.

Let’s express the later terms in terms of the earlier terms by substituting the equations into each other.

From (1), we can write $a = \log_3 4$.
From (2), we can write $b = \log_4 5$.
From (3), we can write $c = \log_5 6$.
From (4), we can write $d = \log_6 7$.
From (5), we can write $e = \log_7 8$.
From (6), we can write $f = \log_8 9$.

Now, we can multiply these logarithmic values together:
$abcdef = (\log_3 4) (\log_4 5) (\log_5 6) (\log_6 7) (\log_7 8) (\log_8 9)$

We can use the change of base formula for logarithms, which states that $\log_x y = \frac{\log_k y}{\log_k x}$ for any valid base $k$. Let’s use base 10 or the natural logarithm for simplicity. However, a more direct way to simplify this product is to observe the telescoping nature of the logarithmic product.

Using the property $\log_x y \cdot \log_y z = \log_x z$:
$(\log_3 4) (\log_4 5) = \log_3 5$
$(\log_3 5) (\log_5 6) = \log_3 6$
$(\log_3 6) (\log_6 7) = \log_3 7$
$(\log_3 7) (\log_7 8) = \log_3 8$
$(\log_3 8) (\log_8 9) = \log_3 9$

So, $abcdef = \log_3 9$.

We know that $3^2 = 9$. Therefore, $\log_3 9 = 2$.

Alternatively, we can substitute the equations sequentially:
Start with $3^a = 4$.
Substitute this into $4^b = 5$: $(3^a)^b = 5 \implies 3^{ab} = 5$.
Substitute this into $5^c = 6$: $(3^{ab})^c = 6 \implies 3^{abc} = 6$.
Substitute this into $6^d = 7$: $(3^{abc})^d = 7 \implies 3^{abcd} = 7$.
Substitute this into $7^e = 8$: $(3^{abcd})^e = 8 \implies 3^{abcde} = 8$.
Substitute this into $8^f = 9$: $(3^{abcde})^f = 9 \implies 3^{abcdef} = 9$.

Since $3^2 = 9$, we can equate the exponents:
$abcdef = 2$.

Final_Answer:2

Ans: C

Explanation:Given the equations:
1) $\log_{x}(x^2 + 12) = 4$
2) $3 \log_{y} x = 1$

From equation (1), using the definition of logarithm, we have:
$x^4 = x^2 + 12$

Rearranging the terms to form a quadratic equation in terms of $x^2$:
$x^4 – x^2 – 12 = 0$

Let $z = x^2$. Then the equation becomes:
$z^2 – z – 12 = 0$

Factor the quadratic equation:
$(z – 4)(z + 3) = 0$

This gives two possible values for $z$: $z = 4$ or $z = -3$.
Since $z = x^2$ and $x$ is a positive real number, $x^2$ must be positive.
Therefore, $x^2 = 4$.
Since $x$ is a positive real number, $x = \sqrt{4} = 2$.

Now, substitute the value of $x$ into equation (2):
$3 \log_{y} 2 = 1$

Divide by 3:
$\log_{y} 2 = \frac{1}{3}$

Using the definition of logarithm:
$y^{\frac{1}{3}} = 2$

To find $y$, cube both sides of the equation:
$(y^{\frac{1}{3}})^3 = 2^3$
$y = 8$

We need to find the value of $x + y$.
$x + y = 2 + 8 = 10$.

Let’s check the conditions for the base of the logarithm.
For $\log_{x}(x^2 + 12)$, the base $x$ must be positive and not equal to 1. We found $x=2$, which satisfies these conditions.
For $\log_{y} x$, the base $y$ must be positive and not equal to 1. We found $y=8$, which satisfies these conditions.

The value of $x + y$ is 10.

Comparing this with the given options:
Option A: 20
Option B: 68
Option C: 10
Option D: 11

The calculated value matches Option C.

The final answer is $\boxed{10}$.
Correct_Option: C

Ans: C

Explanation:Let the given equation be
$\sqrt{5x+9} + \sqrt{5x – 9} = 3(2 + \sqrt{2})$
Square both sides of the equation:
$(\sqrt{5x+9} + \sqrt{5x – 9})^2 = (3(2 + \sqrt{2}))^2$
$(5x+9) + (5x-9) + 2\sqrt{(5x+9)(5x-9)} = 9(4 + 2\sqrt{2} + 2)$
$10x + 2\sqrt{(5x)^2 – 9^2} = 9(6 + 2\sqrt{2})$
$10x + 2\sqrt{25x^2 – 81} = 54 + 18\sqrt{2}$
This approach seems to complicate the problem significantly. Let’s try a different approach.

Let $a = \sqrt{5x+9}$ and $b = \sqrt{5x-9}$.
Then $a^2 = 5x+9$ and $b^2 = 5x-9$.
So $a^2 – b^2 = (5x+9) – (5x-9) = 18$.
We are given $a + b = 3(2 + \sqrt{2})$.
We know that $a^2 – b^2 = (a+b)(a-b)$.
So, $18 = (3(2 + \sqrt{2}))(a-b)$.
$a-b = \frac{18}{3(2 + \sqrt{2})} = \frac{6}{2 + \sqrt{2}}$
Rationalize the denominator:
$a-b = \frac{6(2 – \sqrt{2})}{(2 + \sqrt{2})(2 – \sqrt{2})} = \frac{6(2 – \sqrt{2})}{4 – 2} = \frac{6(2 – \sqrt{2})}{2} = 3(2 – \sqrt{2})$
Now we have two equations:
1) $a + b = 3(2 + \sqrt{2}) = 6 + 3\sqrt{2}$
2) $a – b = 3(2 – \sqrt{2}) = 6 – 3\sqrt{2}$
Add equation (1) and (2):
$(a+b) + (a-b) = (6 + 3\sqrt{2}) + (6 – 3\sqrt{2})$
$2a = 12$
$a = 6$
Substitute $a=6$ into equation (1):
$6 + b = 6 + 3\sqrt{2}$
$b = 3\sqrt{2}$
Now we use the definitions of $a$ and $b$:
$a = \sqrt{5x+9} = 6$
Square both sides:
$5x+9 = 36$
$5x = 36 – 9$
$5x = 27$
$x = \frac{27}{5}$
Let’s check with $b$:
$b = \sqrt{5x-9} = 3\sqrt{2}$
Square both sides:
$5x-9 = (3\sqrt{2})^2 = 9 \times 2 = 18$
$5x = 18 + 9$
$5x = 27$
$x = \frac{27}{5}$
Both equations give the same value for $x$.

We need to find the value of $\sqrt{10x+9}$.
Substitute $x = \frac{27}{5}$:
$10x+9 = 10\left(\frac{27}{5}\right) + 9 = 2 \times 27 + 9 = 54 + 9 = 63$.
So, $\sqrt{10x+9} = \sqrt{63}$.
$\sqrt{63} = \sqrt{9 \times 7} = \sqrt{9} \times \sqrt{7} = 3\sqrt{7}$.

Let’s re-examine the options to ensure our calculations are correct.
Option A: $3\sqrt{31} = \sqrt{9 \times 31} = \sqrt{279}$
Option B: $4\sqrt{5} = \sqrt{16 \times 5} = \sqrt{80}$
Option C: $3\sqrt{7} = \sqrt{9 \times 7} = \sqrt{63}$
Option D: $2\sqrt{7} = \sqrt{4 \times 7} = \sqrt{28}$

Our calculated value is $3\sqrt{7}$.

The final answer is $\boxed{3\sqrt{7}}$.

Correct_Option:C

Ans: D

The provided equation is $2^{4x^{2}}-2^{2x^{2}+x+16}+2^{2x+30}=0$.

This can be rewritten by observing the exponents and terms:
=>$\left(2^{2x^2}\right)^2-2^{2x^2}\cdot2^{x+15}\cdot2^1+\left(2^{x+15}\right)^{^2}=0$

This structure resembles a perfect square trinomial, allowing us to express it as:
=> $\left(2^{2x^2}-2^{x+15}\right)^{^2}=0$

For the square of an expression to be zero, the expression itself must be zero:
=> $2^{2x^2}-2^{x+15}=0$

Now, if two powers with the same base are equal, their exponents must be equal:
=> $2x^2\ =\ x+15$

Rearranging this into a standard quadratic equation:
=> $2x^2-x-15=0$

We can factor this quadratic equation. One way to do this is by splitting the middle term:
=> $2x^2-6x+5x-15=0$

Factoring by grouping:
=> $2x\left(x-3\right)+5\left(x-3\right)=0$

This leads to the factored form:
=> $\left(2x+5\right)\left(x-3\right)\ =\ 0$

The possible values for x are found by setting each factor to zero:
=> $2x+5=0 \implies x = -\frac{5}{2}$
=> $x-3=0 \implies x = 3$

Thus, the possible values of x are $-\frac{5}{2}$ and $3$.

The sum of these possible values is:
=> $\left(3 + (-\frac{5}{2})\right) = 3 – \frac{5}{2} = \frac{6}{2} – \frac{5}{2} = \frac{1}{2}$

The correct option is D.

Ans: 7

**Explanation:**
The given equation is:
$\log_{\sqrt{3}}{(x)}+\frac{\log_{x}{(25)}}{\log_{x}{(0.008)}}=\frac{16}{3}$

First, let’s simplify the terms in the equation.
The first term can be rewritten using the change of base formula for logarithms:
$\log_{\sqrt{3}}{(x)} = \frac{\log_{3}{(x)}}{\log_{3}{(\sqrt{3})}} = \frac{\log_{3}{(x)}}{1/2} = 2\log_{3}{(x)}$

Now, let’s simplify the second term:
$\frac{\log_{x}{(25)}}{\log_{x}{(0.008)}} = \frac{\log_{x}{(5^2)}}{\log_{x}{(8/1000)}} = \frac{2\log_{x}{(5)}}{\log_{x}{(1/125)}} = \frac{2\log_{x}{(5)}}{\log_{x}{(5^{-3})}} = \frac{2\log_{x}{(5)}}{-3\log_{x}{(5)}}$
Assuming $\log_{x}{(5)} \neq 0$, which is true since $x$ is a positive real number and $5 \neq 1$.
So, the second term simplifies to $-\frac{2}{3}$.

Substitute these simplified terms back into the original equation:
$2\log_{3}{(x)} – \frac{2}{3} = \frac{16}{3}$

Now, solve for $\log_{3}{(x)}$:
$2\log_{3}{(x)} = \frac{16}{3} + \frac{2}{3}$
$2\log_{3}{(x)} = \frac{18}{3}$
$2\log_{3}{(x)} = 6$
$\log_{3}{(x)} = 3$

We need to find the value of $\log_{3}({3x^{2}})$.
Using the logarithm properties:
$\log_{3}({3x^{2}}) = \log_{3}{(3)} + \log_{3}{(x^{2})}$
$\log_{3}({3x^{2}}) = 1 + 2\log_{3}{(x)}$

We already found that $\log_{3}{(x)} = 3$. Substitute this value:
$\log_{3}({3x^{2}}) = 1 + 2(3)$
$\log_{3}({3x^{2}}) = 1 + 6$
$\log_{3}({3x^{2}}) = 7$

**Final Answer:**
7

Ans: 5

Explanation:We are given two inequalities:
1) $5^{n-1} < 3^{n+1}$
2) $3^{n+1} < 2^{n+m}$

We need to find the least integer value of $m$ that satisfies the second inequality for all natural numbers $n$ that satisfy the first inequality.

First, let’s analyze the first inequality: $5^{n-1} < 3^{n+1}$.
We can rewrite this as:
$\frac{5^n}{5} < 3^n \cdot 3$
$5^n < 15 \cdot 3^n$
$\frac{5^n}{3^n} < 15$
$(\frac{5}{3})^n < 15$

Let’s test values of $n$ for this inequality:
For $n=1$: $(\frac{5}{3})^1 = \frac{5}{3} \approx 1.67$, which is less than 15. So $n=1$ satisfies the first inequality.
For $n=2$: $(\frac{5}{3})^2 = \frac{25}{9} \approx 2.78$, which is less than 15. So $n=2$ satisfies the first inequality.
For $n=3$: $(\frac{5}{3})^3 = \frac{125}{27} \approx 4.63$, which is less than 15. So $n=3$ satisfies the first inequality.
For $n=4$: $(\frac{5}{3})^4 = \frac{625}{81} \approx 7.72$, which is less than 15. So $n=4$ satisfies the first inequality.
For $n=5$: $(\frac{5}{3})^5 = \frac{3125}{243} \approx 12.86$, which is less than 15. So $n=5$ satisfies the first inequality.
For $n=6$: $(\frac{5}{3})^6 = \frac{15625}{729} \approx 21.43$, which is greater than 15. So $n=6$ does not satisfy the first inequality.

Since $(\frac{5}{3})^n$ is an increasing function, the natural numbers $n$ that satisfy $5^{n-1} < 3^{n+1}$ are $n = 1, 2, 3, 4, 5$.

Now, we need to find the least integer value of $m$ that satisfies $3^{n+1} < 2^{n+m}$ for each of these values of $n$.
We can rewrite the second inequality as:
$3^{n+1} < 2^n \cdot 2^m$
$\frac{3^{n+1}}{2^n} < 2^m$
$3 \cdot \frac{3^n}{2^n} < 2^m$
$3 \cdot (\frac{3}{2})^n < 2^m$

We need to find the maximum value of $3 \cdot (\frac{3}{2})^n$ for $n \in \{1, 2, 3, 4, 5\}$ and then find the least integer $m$ such that $2^m$ is greater than this maximum value.

Let’s calculate $3 \cdot (\frac{3}{2})^n$ for $n=1, 2, 3, 4, 5$:
For $n=1$: $3 \cdot (\frac{3}{2})^1 = 3 \cdot \frac{3}{2} = \frac{9}{2} = 4.5$
For $n=2$: $3 \cdot (\frac{3}{2})^2 = 3 \cdot \frac{9}{4} = \frac{27}{4} = 6.75$
For $n=3$: $3 \cdot (\frac{3}{2})^3 = 3 \cdot \frac{27}{8} = \frac{81}{8} = 10.125$
For $n=4$: $3 \cdot (\frac{3}{2})^4 = 3 \cdot \frac{81}{16} = \frac{243}{16} = 15.1875$
For $n=5$: $3 \cdot (\frac{3}{2})^5 = 3 \cdot \frac{243}{32} = \frac{729}{32} = 22.78125$

The maximum value of $3 \cdot (\frac{3}{2})^n$ for $n \in \{1, 2, 3, 4, 5\}$ is $22.78125$.
We need $2^m > 22.78125$.

Let’s check powers of 2:
$2^1 = 2$
$2^2 = 4$
$2^3 = 8$
$2^4 = 16$
$2^5 = 32$

The smallest power of 2 that is greater than $22.78125$ is $32$, which corresponds to $m=5$.

Therefore, the least integer value of $m$ is 5.

Final_Answer:5

Ans: 47

Explanation:The inequality is given by $\frac{4-\log_{2}n}{3-\log_{4}n} < 0$.
For the logarithm to be defined, we must have $n > 0$.

We can rewrite $\log_{4}n$ in terms of $\log_{2}n$:
$\log_{4}n = \frac{\log_{2}n}{\log_{2}4} = \frac{\log_{2}n}{2}$.

Substitute this into the inequality:
$\frac{4-\log_{2}n}{3-\frac{\log_{2}n}{2}} < 0$

Multiply the numerator and denominator of the denominator by 2 to simplify:
$\frac{4-\log_{2}n}{\frac{6-\log_{2}n}{2}} < 0$
$ \frac{2(4-\log_{2}n)}{6-\log_{2}n} < 0$
$ \frac{8-2\log_{2}n}{6-\log_{2}n} < 0$

Let $x = \log_{2}n$. The inequality becomes:
$\frac{8-2x}{6-x} < 0$

To solve this inequality, we consider the signs of the numerator and the denominator.
Case 1: Numerator is positive and denominator is negative.
$8-2x > 0 \implies 8 > 2x \implies 4 > x$
$6-x < 0 \implies 6 < x$
This case gives $x < 4$ and $x > 6$, which is impossible.

Case 2: Numerator is negative and denominator is positive.
$8-2x < 0 \implies 8 < 2x \implies 4 < x$
$6-x > 0 \implies 6 > x$
This case gives $x > 4$ and $x < 6$, so $4 < x < 6$.

Now substitute back $x = \log_{2}n$:
$4 < \log_{2}n < 6$

Convert the logarithms to exponential form:
$2^4 < n < 2^6$
$16 < n < 64$

We also need to consider the condition that the denominator cannot be zero.
$3-\log_{4}n \neq 0$
$\log_{4}n \neq 3$
$n \neq 4^3$
$n \neq 64$
Our inequality $16 < n < 64$ already excludes $n=64$.

So, the values of $n$ must be integers strictly between 16 and 64.
The integers are $17, 18, \ldots, 63$.

To find the number of distinct integer values, we use the formula: Last – First + 1.
Number of values = $63 – 17 + 1 = 46 + 1 = 47$.

The number of distinct integer values of $n$ satisfying the inequality is 47.

Final Answer: The number of distinct integer values of n satisfying the inequality is 47.
The range of integer values for n is from 17 to 63 inclusive.
The number of integers in this range is $63 – 17 + 1 = 47$.

Final_Answer:47

Ans: 14

Given the equation:
$(\sqrt{\frac{7}{5}})^{3x-y}=\frac{875}{2401}$

We can rewrite the left side using exponent rules:
$\left(\frac{7}{5}\right)^{\frac{\left(3x-y\right)}{2}}=\frac{875}{2401}$

Now, let’s express the right side with a base of $\frac{7}{5}$. We observe that $875 = 5^3 \times 7$ and $2401 = 7^4$.
So, $\frac{875}{2401} = \frac{5^3 \times 7}{7^4} = \frac{5^3}{7^3} = \left(\frac{5}{7}\right)^3 = \left(\frac{7}{5}\right)^{-3}$.

Substituting this back into the equation:
$\left(\frac{7}{5}\right)^{\frac{\left(3x-y\right)}{2}}=\left(\frac{7}{5}\right)^{-3}$

For the equality to hold, the exponents must be equal:
$\frac{3x-y}{2} = -3$
$3x-y = -6 \quad (*)$

Now consider the second equation:
$(\frac{4a}{b})^{6x-y}=(\frac{2a}{b})^{y-6x}$

We can rewrite the right side as:
$(\frac{2a}{b})^{y-6x} = \left(\left(\frac{2a}{b}\right)^{-1}\right)^{6x-y} = \left(\frac{b}{2a}\right)^{6x-y}$

The equation becomes:
$(\frac{4a}{b})^{6x-y}=\left(\frac{b}{2a}\right)^{6x-y}$

Since the bases are not necessarily equal, for the equality to hold, the exponent must be zero:
$6x-y = 0$
$y = 6x$

Now we have a system of two linear equations:
1) $3x-y = -6$
2) $y = 6x$

Substitute equation (2) into equation (1):
$3x – (6x) = -6$
$-3x = -6$
$x = 2$

Now substitute the value of $x$ back into equation (2):
$y = 6 \times 2$
$y = 12$

We need to find $x+y$:
$x+y = 2+12 = 14$

The final answer is $\boxed{14}$.

Ans: D

Explanation:Given that $(3+2\sqrt{2})$ is a root of the equation $ax^{2}+bx+c=0$. Since the coefficients a, b, and c are integers, if one root is of the form $p + q\sqrt{r}$, then its conjugate $p – q\sqrt{r}$ must also be a root.
So, the roots of $ax^{2}+bx+c=0$ are $(3+2\sqrt{2})$ and $(3-2\sqrt{2})$.
Using Vieta’s formulas for this equation:
Sum of roots: $(3+2\sqrt{2}) + (3-2\sqrt{2}) = 6$
$-b/a = 6 \Rightarrow b = -6a$
Product of roots: $(3+2\sqrt{2})(3-2\sqrt{2}) = 3^2 – (2\sqrt{2})^2 = 9 – (4 \times 2) = 9 – 8 = 1$
$c/a = 1 \Rightarrow c = a$

Given that $(4+2\sqrt{3})$ is a root of the equation $ay^{2}+my+n=0$. Since the coefficients a, m, and n are integers, if one root is of the form $p + q\sqrt{r}$, then its conjugate $p – q\sqrt{r}$ must also be a root.
So, the roots of $ay^{2}+my+n=0$ are $(4+2\sqrt{3})$ and $(4-2\sqrt{3})$.
Using Vieta’s formulas for this equation:
Sum of roots: $(4+2\sqrt{3}) + (4-2\sqrt{3}) = 8$
$-m/a = 8 \Rightarrow m = -8a$
Product of roots: $(4+2\sqrt{3})(4-2\sqrt{3}) = 4^2 – (2\sqrt{3})^2 = 16 – (4 \times 3) = 16 – 12 = 4$
$n/a = 4 \Rightarrow n = 4a$

We need to find the value of $(\frac{b}{m}+\frac{c-2b}{n})$.
Substitute the values of b, c, m, and n in terms of a:
$b = -6a$
$c = a$
$m = -8a$
$n = 4a$

$\frac{b}{m} = \frac{-6a}{-8a} = \frac{6}{8} = \frac{3}{4}$

$c-2b = a – 2(-6a) = a + 12a = 13a$

$\frac{c-2b}{n} = \frac{13a}{4a} = \frac{13}{4}$

$(\frac{b}{m}+\frac{c-2b}{n}) = \frac{3}{4} + \frac{13}{4} = \frac{3+13}{4} = \frac{16}{4} = 4$

Correct_Option:D

Ans: 99

Explanation:The given equation is:
$5 – \log_{10}\sqrt{1 + x} + 4 \log_{10} \sqrt{1 – x} = \log_{10} \frac{1}{\sqrt{1 – x^2}}$

We can rewrite the equation using logarithm properties:
$5 – \frac{1}{2}\log_{10}(1 + x) + 4 \cdot \frac{1}{2}\log_{10}(1 – x) = \log_{10}((1 – x^2)^{-1/2})$
$5 – \frac{1}{2}\log_{10}(1 + x) + 2\log_{10}(1 – x) = -\frac{1}{2}\log_{10}(1 – x^2)$

We know that $1 – x^2 = (1 – x)(1 + x)$. So, $\log_{10}(1 – x^2) = \log_{10}((1 – x)(1 + x)) = \log_{10}(1 – x) + \log_{10}(1 + x)$.

Substitute this into the equation:
$5 – \frac{1}{2}\log_{10}(1 + x) + 2\log_{10}(1 – x) = -\frac{1}{2}(\log_{10}(1 – x) + \log_{10}(1 + x))$
$5 – \frac{1}{2}\log_{10}(1 + x) + 2\log_{10}(1 – x) = -\frac{1}{2}\log_{10}(1 – x) – \frac{1}{2}\log_{10}(1 + x)$

Now, let’s move all the logarithm terms to one side and the constant to the other:
$5 = \frac{1}{2}\log_{10}(1 + x) – \frac{1}{2}\log_{10}(1 + x) – 2\log_{10}(1 – x) – \frac{1}{2}\log_{10}(1 – x)$
$5 = 0 \cdot \log_{10}(1 + x) + (-\frac{4}{2} – \frac{1}{2})\log_{10}(1 – x)$
$5 = -\frac{5}{2}\log_{10}(1 – x)$

Now, solve for $\log_{10}(1 – x)$:
$\log_{10}(1 – x) = 5 \cdot (-\frac{2}{5})$
$\log_{10}(1 – x) = -2$

Convert this logarithmic equation to an exponential equation:
$1 – x = 10^{-2}$
$1 – x = \frac{1}{100}$

Solve for x:
$x = 1 – \frac{1}{100}$
$x = \frac{100 – 1}{100}$
$x = \frac{99}{100}$

The question asks for the value of 100x:
$100x = 100 \cdot \frac{99}{100}$
$100x = 99$

We also need to ensure that the terms in the logarithm are positive for the expression to be defined.
$1+x > 0 \implies x > -1$
$1-x > 0 \implies x < 1$
$1-x^2 > 0 \implies -1 < x < 1$
Our obtained value of $x = 99/100$ satisfies these conditions.

Final_Answer:99

Ans: 7

We are given the inequality $2.25\leq2+2^{n+2}\leq202$.
Subtracting 2 from all parts of the inequality, we get:
$2.25-2\le2+2^{n+2}-2\le202-2$
$0.25\le2^{n+2}\le200$

Taking the base-2 logarithm of all parts:
$\log_20.25\le n+2\le\log_2200$

We know that $\log_20.25 = \log_2(1/4) = \log_2(2^{-2}) = -2$.
Also, $\log_2200$ is between $\log_2128=7$ and $\log_2256=8$. Let’s approximate it as $7.xx$.
So, the inequality becomes:
$-2\le n+2\le7.xx$

Subtracting 2 from all parts:
$-2-2\le n+2-2\le7.xx-2$
$-4\le n\le5.xx$

The possible integer values for $n$ from this inequality are: -4, -3, -2, -1, 0, 1, 2, 3, 4, 5.

Now consider the second expression provided, $3+3^{n+1}$. For this expression to yield an integer value, the exponent $n+1$ must be such that $3^{n+1}$ is defined and results in a value that, when added to 3, gives an integer. This implies that $n+1$ should not lead to a fractional or non-integer result for $3^{n+1}$ in the context of typical integer arithmetic problems. If we assume $n$ is an integer, then $n+1$ is also an integer. For $3^{n+1}$ to be a standard integer, $n+1 \ge 0$ is typically implied if we are looking for positive integer powers or $n+1$ can be any integer. However, if the context requires the result of $3^{n+1}$ to be an integer that combines with 3 to form a valid number in the problem set, we should consider values of $n$ that are at least -1 to ensure $n+1 \ge 0$, which makes $3^{n+1}$ an integer (e.g., $3^0=1$, $3^1=3$, etc.). If $n+1$ were negative, say $n=-2$, then $n+1 = -1$, and $3^{-1} = 1/3$, which is not an integer. Therefore, for the second expression to be an integer, we require $n+1 \ge 0$, which means $n \ge -1$.

Combining this condition with the possible integer values of $n$ from the first inequality, the common integer values are: -1, 0, 1, 2, 3, 4, 5.

There are a total of 7 possible integer values for $n$.

Ans: B

Explanation:
The given quadratic equation is $ax^2 – bx + c = 0$, where a, b, and c are rational numbers and $a \neq 0$.
One of the roots is given as $2+\sqrt{3}$.
Since the coefficients of the quadratic equation are rational, if one root is of the form $p+\sqrt{q}$ where p and q are rational and $\sqrt{q}$ is irrational, then the conjugate $p-\sqrt{q}$ must also be a root.
Therefore, the other root of the equation is $2-\sqrt{3}$.

Let the roots of the equation be $\alpha$ and $\beta$.
So, $\alpha = 2+\sqrt{3}$ and $\beta = 2-\sqrt{3}$.

From Vieta’s formulas, we have:
Sum of roots: $\alpha + \beta = -\frac{-b}{a} = \frac{b}{a}$
Product of roots: $\alpha \beta = \frac{c}{a}$

Calculate the sum of the roots:
$\alpha + \beta = (2+\sqrt{3}) + (2-\sqrt{3}) = 2+2+\sqrt{3}-\sqrt{3} = 4$
So, $\frac{b}{a} = 4$. This implies $b = 4a$.

Calculate the product of the roots:
$\alpha \beta = (2+\sqrt{3})(2-\sqrt{3})$
Using the difference of squares formula $(x+y)(x-y) = x^2 – y^2$:
$\alpha \beta = 2^2 – (\sqrt{3})^2 = 4 – 3 = 1$
So, $\frac{c}{a} = 1$. This implies $c = a$.

We are also given the condition $b = c^3$.
Substitute the relationships we found:
$4a = a^3$

Since $a \neq 0$, we can divide both sides by $a$:
$4 = a^2$

Taking the square root of both sides:
$a = \pm \sqrt{4}$
$a = \pm 2$

We need to find $|a|$.
$|a| = |2| = 2$ or $|a| = |-2| = 2$.
In either case, $|a| = 2$.

Let’s verify if this is consistent.
If $a=2$, then $b=4a = 4(2)=8$ and $c=a=2$.
Check the condition $b=c^3$: $8 = 2^3$, which is true.
The equation is $2x^2 – 8x + 2 = 0$, or $x^2 – 4x + 1 = 0$.
Roots are $\frac{-(-4) \pm \sqrt{(-4)^2 – 4(1)(1)}}{2(1)} = \frac{4 \pm \sqrt{16-4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$. This matches the given root.

If $a=-2$, then $b=4a = 4(-2)=-8$ and $c=a=-2$.
Check the condition $b=c^3$: $-8 = (-2)^3$, which is true.
The equation is $-2x^2 – (-8)x + (-2) = 0$, or $-2x^2 + 8x – 2 = 0$, or $x^2 – 4x + 1 = 0$.
The roots are again $2 \pm \sqrt{3}$.

Therefore, $|a|=2$.

Correct_Option: B

Ans: 5

We are presented with the equation:
$\log_2\left\{3+\log_3\left\{4+\log_4\left(x-1\right)\right\}\right\}=2$
Applying the definition of logarithms, we convert the outermost logarithm. The base is 2 and the result of the logarithm is 2, so the argument must be $2^2 = 4$. This yields:
$3+\log_3\left\{4+\log_4\left(x-1\right)\right\}=4$
Next, we isolate the $\log_3$ term by subtracting 3 from both sides:
$\log_3\left\{4+\log_4\left(x-1\right)\right\}=1$
Again, using the definition of logarithms, with a base of 3 and a result of 1, the argument must be $3^1 = 3$:
$4+\log_4\left(x-1\right)=3$
Now, we isolate the $\log_4$ term by subtracting 4 from both sides:
$\log_4\left(x-1\right)=-1$
Finally, we apply the definition of logarithms one last time. With a base of 4 and a result of -1, the argument must be $4^{-1}$:
$x-1 = 4^{-1}$
$x-1 = \frac{1}{4}$
To solve for $x$, we add 1 to both sides:
$x = \frac{1}{4} + 1$
$x = \frac{1}{4} + \frac{4}{4}$
$x = \frac{5}{4}$
Multiplying both sides by 4 to eliminate the fraction gives:
$4x = 5$

Ans: C

Let the given equation be:
$\frac{\log_{15}{a}+\log_{32}{a}}{(\log_{15}{a})(\log_{32}{a})}=4$

Using the change of base formula for logarithms ($\log_b x = \frac{\log x}{\log b}$), we can rewrite the terms:
$\frac{\left(\frac{\log a}{\log 15}+\frac{\log a}{\log 32}\right)}{\frac{\log a}{\log 15}\times\ \frac{\log a}{\log32}\ \ }=4$

Now, let’s simplify the numerator by factoring out $\log a$:
$\frac{\log a\left(\frac{1}{\log 15}+\frac{1}{\log 32}\right)}{\frac{(\log a)^2}{(\log 15)(\log 32)}\ \ }=4$

Combine the fractions in the numerator:
$\frac{\log a\left(\frac{\log 32+\log 15}{(\log 15)(\log 32)}\right)}{\frac{(\log a)^2}{(\log 15)(\log 32)}\ \ }=4$

Now, we can cancel out the common term $(\log 15)(\log 32)$ from the numerator and the denominator of the main fraction:
$\frac{\log a(\log 32+\log 15)}{(\log a)^2}=4$

Simplify by canceling one $\log a$ term:
$\frac{\log 32+\log 15}{\log a}=4$

Rearrange the terms:
$\log 32+\log 15 = 4\log a$

Using the logarithm property $\log x + \log y = \log(xy)$:
$\log (32 \times 15) = \log a^4$
$\log 480 = \log a^4$

Since the logarithms are equal, their arguments must be equal:
$a^4 = 480$

To estimate the value of $a$, we can consider powers of integers.
$4^4 = 256$
$5^4 = 625$

Since $256 < 480 < 625$, we can conclude that $a$ lies between 4 and 5.

Ans: 6

The problem presents an inequality: $(\sqrt[7]{10})(\sqrt[7]{10})^{2}…(\sqrt[7]{10})^{n}>999$.

This can be simplified by combining the terms with the same base: $(\sqrt[7]{10})^{1+2+…+n}>999$.

Next, we can rewrite the expression using exponent notation: $10^{\frac{1+2+…+n}{7}}>999$.

To find the minimum integer value of $n$ that satisfies this condition, we consider the smallest integer exponent for base 10 that would exceed 999. Since $10^3 = 1000$, which is greater than 999, the smallest integer exponent required is 3.

Therefore, we set the exponent in our inequality equal to 3: $\frac{1+2+…+n}{7}=3$.

This simplifies to the sum of the first $n$ integers being equal to 21: $1+2+…+n = 21$.

We know that the sum of the first $n$ positive integers is given by the formula $\frac{n(n+1)}{2}$. However, in this specific problem, we can deduce the value of $n$ by inspection. By summing consecutive integers starting from 1, we find that when $n=6$, the sum is $1+2+3+4+5+6 = 21$.

Thus, the minimum integer value of $n$ is 6.

Ans: B

The provided equation is:
$2^{Y^2({\log_{3}{5})}}=5^{Y^2(\log_3 2)}$

We are also given that:
$5^{Y^2\left(\log_32\right)}=5^{\left(\log_23\right)}$

From the second given equality, we can equate the exponents:
$Y^2\left(\log_32\right)=\left(\log_23\right)$

Now, we can isolate $Y^2$:
$Y^2=\frac{\log_23}{\log_32}$

Using the change of base formula for logarithms, which states $\log_b a = \frac{1}{\log_a b}$, we can rewrite $\frac{1}{\log_32}$ as $\log_23$. Therefore:
$Y^2 = (\log_23) \times (\log_23)$
$Y^2 = (\log_23)^2$

Taking the square root of both sides, we get two possible values for Y:
$Y = (\log_23) \text{ or } Y = -(\log_23)$

The problem statement indicates that Y is a negative number. Therefore, we select the negative solution:
$Y = -(\log_23)$

Using the property of logarithms that $-\log_b a = \log_b (a^{-1}) = \log_b \left(\frac{1}{a}\right)$, we can rewrite the solution as:
$Y = \log_2\left(\frac{1}{3}\right)$

Ans: C

Given is the value of $x = 2^{12(7+4\sqrt{3})}$.

We are asked to find the value of $\frac{x^{7/2}}{x^{2\sqrt{3}}}$.

Using the properties of exponents, we can simplify this expression:

$ \frac{x^{7/2}}{x^{2\sqrt{3}}} = x^{\frac{7}{2} – 2\sqrt{3}} $

First, let’s express the exponents in terms of the base 2.

From the given $x=2^{12(7+4\sqrt{3})}$, we have:
$ x^{7/2} = (2^{12(7+4\sqrt{3})})^{7/2} = 2^{12(7+4\sqrt{3}) \times \frac{7}{2}} = 2^{6 \times 7 (7+4\sqrt{3})} = 2^{42(7+4\sqrt{3})} $

And
$ x^{2\sqrt{3}} = (2^{12(7+4\sqrt{3})})^{2\sqrt{3}} = 2^{12(7+4\sqrt{3}) \times 2\sqrt{3}} = 2^{24\sqrt{3}(7+4\sqrt{3})} $

Now, let’s divide these two expressions:
$ \frac{x^{7/2}}{x^{2\sqrt{3}}} = \frac{2^{42(7+4\sqrt{3})}}{2^{24\sqrt{3}(7+4\sqrt{3})}} = 2^{42(7+4\sqrt{3}) – 24\sqrt{3}(7+4\sqrt{3})} $

We can factor out $(7+4\sqrt{3})$ from the exponent:
$ 2^{(7+4\sqrt{3})(42 – 24\sqrt{3})} $

Let’s simplify the term $(42 – 24\sqrt{3})$. We can factor out 6 from this term:
$ 42 – 24\sqrt{3} = 6(7 – 4\sqrt{3}) $

Substitute this back into the exponent:
$ 2^{(7+4\sqrt{3}) \times 6(7 – 4\sqrt{3})} $

Rearranging the terms, we get:
$ 2^{6 \times (7+4\sqrt{3})(7 – 4\sqrt{3})} $

Now, we recognize the form $(a+b)(a-b) = a^2 – b^2$. Here, $a=7$ and $b=4\sqrt{3}$.
So, $(7+4\sqrt{3})(7 – 4\sqrt{3}) = 7^2 – (4\sqrt{3})^2 = 49 – (16 \times 3) = 49 – 48 = 1$.

Substituting this back into the expression:
$ 2^{6 \times 1} = 2^6 $

Thus, the value of the expression is $2^6$.

Therefore, option C is the correct choice.

Ans: 36

Explanation:We are given the equation $\log_{4}{5}=(\log_{4}{y})(\log_{6}{\sqrt{5}})$.
We can rewrite $\log_{6}{\sqrt{5}}$ using the property $\log_{a}{b^c} = c\log_{a}{b}$:
$\log_{6}{\sqrt{5}} = \log_{6}{5^{1/2}} = \frac{1}{2}\log_{6}{5}$.

Now substitute this back into the original equation:
$\log_{4}{5} = (\log_{4}{y})\left(\frac{1}{2}\log_{6}{5}\right)$

We can use the change of base formula for logarithms, $\log_{a}{b} = \frac{\log_{c}{b}}{\log_{c}{a}}$. Let’s change the base of $\log_{6}{5}$ to base 4:
$\log_{6}{5} = \frac{\log_{4}{5}}{\log_{4}{6}}$.

Substitute this into the equation:
$\log_{4}{5} = (\log_{4}{y})\left(\frac{1}{2}\frac{\log_{4}{5}}{\log_{4}{6}}\right)$

We can divide both sides by $\log_{4}{5}$ (since $\log_{4}{5} \neq 0$):
$1 = (\log_{4}{y})\left(\frac{1}{2\log_{4}{6}}\right)$

Now, isolate $\log_{4}{y}$:
$\log_{4}{y} = \frac{1}{\frac{1}{2\log_{4}{6}}}$
$\log_{4}{y} = 2\log_{4}{6}$

Using the property $c\log_{a}{b} = \log_{a}{b^c}$:
$\log_{4}{y} = \log_{4}{6^2}$
$\log_{4}{y} = \log_{4}{36}$

Since the logarithms have the same base, the arguments must be equal:
$y = 36$

Alternatively, we could use the property that $\log_b a \cdot \log_c b = \log_c a$.
Let’s rewrite the given equation:
$\log_{4}{5} = \log_{4}{y} \cdot \log_{6}{\sqrt{5}}$
$\log_{4}{5} = \log_{4}{y} \cdot \frac{1}{2}\log_{6}{5}$

Let’s change the base of $\log_{6}{5}$ to base 4:
$\log_{6}{5} = \frac{\log_{4}{5}}{\log_{4}{6}}$

So, the equation becomes:
$\log_{4}{5} = \log_{4}{y} \cdot \frac{1}{2} \cdot \frac{\log_{4}{5}}{\log_{4}{6}}$

Divide both sides by $\log_{4}{5}$:
$1 = \log_{4}{y} \cdot \frac{1}{2} \cdot \frac{1}{\log_{4}{6}}$
$2 \log_{4}{6} = \log_{4}{y}$
$\log_{4}{6^2} = \log_{4}{y}$
$\log_{4}{36} = \log_{4}{y}$
$y = 36$

Final_Answer:36

Ans: C

Upon simplifying the given expression, we arrive at:
$2-\left(\log_ab+\frac{1}{\log_ab}\right)$
Utilizing the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we know that for any positive number $x$, $x + \frac{1}{x} \ge 2$. In this case, let $x = \log_ab$. Therefore, we have:
$\frac{\left(\log_ab+\frac{1}{\log_ab}\right)}{2}\ge1$
Which implies:
$\left(\log_ab+\frac{1}{\log_ab}\right)\ge2$
Substituting this back into the simplified expression, we get:
$2 – (\text{a value greater than or equal to 2})$
This means the expression will always be less than or equal to 0. Consequently, a positive value for the expression is not achievable. Therefore, 1 cannot be the value.

Ans: A

The initial relationships can be expressed as:
$\log_a(30) = A \quad \text{or} \quad \log_a(5) + \log_a(2) + \log_a(3) = A$………(1)
$\log_a\left(\frac{5}{3}\right) = -B \quad \text{or} \quad \log_a(3) – \log_a(5) = B$………(2)

We are also given that:
$\log_a(2) = 3$

Substituting the value of $\log_a(2)$ into equation (1), we obtain:
$\log_a(5) + 3 + \log_a(3) = A$
$\log_a(5) + \log_a(3) = A – 3$

Now we have a system of two equations with two unknowns, $\log_a(3)$ and $\log_a(5)$. Solving this system derived from (1) and (2) yields:
$\log_a(3) = \frac{A+B-3}{2} \quad \text{or} \quad \log_3(a) = \frac{2}{A+B-3}$

Ans: 3

Given that $14^a = 36^b = 84^c = k$.
This implies that $a = \log_{14}k$, $b = \log_{36}k$, and $c = \log_{84}k$.

We are asked to evaluate $6b\left(\frac{1}{c} – \frac{1}{a}\right)$.

First, let’s express $\frac{1}{c}$ and $\frac{1}{a}$ in terms of logarithms with base $k$.
From $c = \log_{84}k$, we get $\frac{1}{c} = \log_k84$.
From $a = \log_{14}k$, we get $\frac{1}{a} = \log_k14$.

Now, substitute these into the expression:
$6b\left(\frac{1}{c} – \frac{1}{a}\right) = 6 \cdot \log_{36}k \cdot (\log_k84 – \log_k14)$.

Using the logarithm property $\log_b x – \log_b y = \log_b \frac{x}{y}$, we have:
$\log_k84 – \log_k14 = \log_k\left(\frac{84}{14}\right) = \log_k6$.

So the expression becomes:
$6 \cdot \log_{36}k \cdot \log_k6$.

We can use the change of base formula for logarithms, which states that $\log_b a = \frac{\log_c a}{\log_c b}$. A useful form of this is $\log_b a \cdot \log_a c = \log_b c$.
Alternatively, we can use the property $\log_b a = \frac{1}{\log_a b}$.
So, $\log_{36}k = \frac{1}{\log_k36}$.

The expression is now:
$6 \cdot \frac{1}{\log_k36} \cdot \log_k6$.

We know that $36 = 6^2$. Therefore, $\log_k36 = \log_k(6^2) = 2 \log_k6$.

Substituting this back into the expression:
$6 \cdot \frac{1}{2 \log_k6} \cdot \log_k6$.

The $\log_k6$ terms cancel out:
$6 \cdot \frac{1}{2} = 3$.

Thus, $6b\left(\frac{1}{c} – \frac{1}{a}\right) = 3$.

Ans: 24

$ \frac{2 \times 4 \times 8 \times 16}{(\log_{2}{4})^2 (\log_{4}{8})^3 (\log_{8}{16})^4} $

First, express all numbers in the numerator as powers of 2:
$ 2 = 2^1 $
$ 4 = 2^2 $
$ 8 = 2^3 $
$ 16 = 2^4 $
So, the numerator becomes:
$ 2^1 \times 2^2 \times 2^3 \times 2^4 = 2^{1+2+3+4} = 2^{10} $

Next, evaluate the logarithmic terms in the denominator. We will use the change of base formula for logarithms, $\log_b a = \frac{\log_c a}{\log_c b}$, and the property $\log_{b^m} b^n = \frac{n}{m}$.

For the first term, $(\log_2 4)^2$:
$ \log_2 4 = \log_2 2^2 = 2 $
So, $(\log_2 4)^2 = 2^2 = 4$.

For the second term, $(\log_4 8)^3$:
$ \log_4 8 = \log_{2^2} 2^3 = \frac{3}{2} $
So, $(\log_4 8)^3 = \left(\frac{3}{2}\right)^3$.

For the third term, $(\log_8 16)^4$:
$ \log_8 16 = \log_{2^3} 2^4 = \frac{4}{3} $
So, $(\log_8 16)^4 = \left(\frac{4}{3}\right)^4$.

Now substitute these back into the original expression:
$ \frac{2^{10}}{(\log_2 4)^2 (\log_4 8)^3 (\log_8 16)^4} = \frac{2^{10}}{(2)^2 \left(\frac{3}{2}\right)^3 \left(\frac{4}{3}\right)^4} $

This simplifies to:
$ \frac{2^{10}}{4 \times \frac{3^3}{2^3} \times \frac{4^4}{3^4}} $

Let’s re-examine the intermediate step from the provided example, which might be slightly different in its simplification path:
$ \frac{2^{10}}{(2)^2 (\frac{3}{2})^3 (\frac{4}{3})^4} $
The example seems to have an error in the last step’s calculation. Let’s proceed with a correct calculation.

$ \frac{2^{10}}{4 \times \frac{27}{8} \times \frac{256}{81}} $

Now, let’s combine the terms in the denominator:
$ 4 \times \frac{27}{8} \times \frac{256}{81} = \frac{4 \times 27 \times 256}{8 \times 81} $
Simplify by cancelling common factors:
$ \frac{1 \times 27 \times 256}{2 \times 81} = \frac{27 \times 256}{162} $
Further simplification:
$ \frac{27 \times 256}{162} = \frac{1 \times 256}{6} = \frac{128}{3} $

So the expression becomes:
$ \frac{2^{10}}{\frac{128}{3}} = \frac{1024}{\frac{128}{3}} = 1024 \times \frac{3}{128} $
$ = 8 \times 3 = 24 $

The original expression evaluates to 24.

Ans: A

Explanation:We are given that $(5.55)^x = 1000$ and $(0.555)^y = 1000$.
From the first equation, we can take the logarithm base 10 on both sides:
$log_{10}(5.55)^x = log_{10}(1000)$
$x \cdot log_{10}(5.55) = 3$
$\frac{1}{x} = \frac{log_{10}(5.55)}{3}$

From the second equation, we can take the logarithm base 10 on both sides:
$log_{10}(0.555)^y = log_{10}(1000)$
$y \cdot log_{10}(0.555) = 3$
$\frac{1}{y} = \frac{log_{10}(0.555)}{3}$

We need to find the value of $\frac{1}{x} – \frac{1}{y}$.
$\frac{1}{x} – \frac{1}{y} = \frac{log_{10}(5.55)}{3} – \frac{log_{10}(0.555)}{3}$
$\frac{1}{x} – \frac{1}{y} = \frac{1}{3} (log_{10}(5.55) – log_{10}(0.555))$
Using the logarithm property $log(a) – log(b) = log(\frac{a}{b})$:
$\frac{1}{x} – \frac{1}{y} = \frac{1}{3} log_{10}\left(\frac{5.55}{0.555}\right)$
$\frac{1}{x} – \frac{1}{y} = \frac{1}{3} log_{10}\left(\frac{5.55}{\frac{555}{1000}}\right)$
$\frac{1}{x} – \frac{1}{y} = \frac{1}{3} log_{10}\left(\frac{555}{100} \times \frac{1000}{555}\right)$
$\frac{1}{x} – \frac{1}{y} = \frac{1}{3} log_{10}\left(\frac{1000}{100}\right)$
$\frac{1}{x} – \frac{1}{y} = \frac{1}{3} log_{10}(10)$
Since $log_{10}(10) = 1$:
$\frac{1}{x} – \frac{1}{y} = \frac{1}{3} \times 1$
$\frac{1}{x} – \frac{1}{y} = \frac{1}{3}$

Alternatively, we can write:
$(5.55)^x = 1000 \implies 5.55 = 1000^{1/x}$
$(0.555)^y = 1000 \implies 0.555 = 1000^{1/y}$
We know that $5.55 = 10 \times 0.555$.
So, $1000^{1/x} = 10 \times 1000^{1/y}$
$1000^{1/x} = 10^1 \times 1000^{1/y}$
$1000^{1/x} = 1000^{1/3} \times 1000^{1/y}$
$1000^{1/x} = 1000^{1/3 + 1/y}$
Equating the exponents:
$\frac{1}{x} = \frac{1}{3} + \frac{1}{y}$
$\frac{1}{x} – \frac{1}{y} = \frac{1}{3}$

Correct_Option:A

Ans: C

We are presented with the equation:
$(\surd2)^{19} 3^4 4^2 9^m 8^n = 3^n 16^m (\sqrt[4]{64})$
To solve this, we will express all terms as powers of their prime factors, specifically 2 and 3.

The left side of the equation becomes:
$2^{\frac{19}{2}} \cdot 3^4 \cdot (2^2)^2 \cdot (3^2)^m \cdot (2^3)^n$
$2^{\frac{19}{2}} \cdot 3^4 \cdot 2^4 \cdot 3^{2m} \cdot 2^{3n}$
Combining the powers of the same base:
$2^{\frac{19}{2} + 4 + 3n} \cdot 3^{4 + 2m}$

The right side of the equation becomes:
$3^n \cdot (2^4)^m \cdot (2^6)^{\frac{1}{4}}$
$3^n \cdot 2^{4m} \cdot 2^{\frac{6}{4}}$
Combining the powers of the same base:
$2^{4m + \frac{6}{4}} \cdot 3^n$

Now, we equate the expressions for both sides:
$2^{\frac{19}{2} + 4 + 3n} \cdot 3^{4 + 2m} = 2^{4m + \frac{6}{4}} \cdot 3^n$

By comparing the exponents of the base 2 on both sides, we obtain the first equation:
$\frac{19}{2} + 4 + 3n = 4m + \frac{6}{4}$
Multiplying by 4 to clear denominators:
$38 + 16 + 12n = 16m + 6$
$54 + 12n = 16m + 6$
$16m – 12n = 48$
Dividing by 4:
$4m – 3n = 12$
$4m = 3n + 12 \quad \cdots (1)$

By comparing the exponents of the base 3 on both sides, we obtain the second equation:
$4 + 2m = n \quad \cdots (2)$

Now, we substitute the value of $n$ from equation (2) into equation (1):
$4m = 3(4 + 2m) + 12$
$4m = 12 + 6m + 12$
$4m = 6m + 24$
$4m – 6m = 24$
$-2m = 24$
$m = -12$

Ans: A

Explanation:We are given two equations:
1) $\log_{5}{(x + y)} + \log_{5}{(x – y)} = 3$
2) $\log_{2}{y} – \log_{2}{x} = 1 – \log_{2}{3}$

Let’s simplify the first equation. Using the logarithm property $\log a + \log b = \log (ab)$:
$\log_{5}{((x + y)(x – y))} = 3$
Using the property $(a+b)(a-b) = a^2 – b^2$:
$\log_{5}{(x^2 – y^2)} = 3$
Convert the logarithmic equation to an exponential equation. If $\log_b a = c$, then $b^c = a$:
$x^2 – y^2 = 5^3$
$x^2 – y^2 = 125$

Now let’s simplify the second equation. Using the logarithm property $\log a – \log b = \log (a/b)$:
$\log_{2}{\frac{y}{x}} = 1 – \log_{2}{3}$
We can rewrite $1$ as $\log_{2}{2}$:
$\log_{2}{\frac{y}{x}} = \log_{2}{2} – \log_{2}{3}$
Using the property $\log a – \log b = \log (a/b)$ again:
$\log_{2}{\frac{y}{x}} = \log_{2}{\frac{2}{3}}$
Since the logarithms are equal and the base is the same, the arguments must be equal:
$\frac{y}{x} = \frac{2}{3}$
From this, we can express y in terms of x:
$y = \frac{2}{3}x$

Now substitute this expression for y into the simplified first equation ($x^2 – y^2 = 125$):
$x^2 – (\frac{2}{3}x)^2 = 125$
$x^2 – \frac{4}{9}x^2 = 125$
Combine the x^2 terms:
$(\frac{9}{9} – \frac{4}{9})x^2 = 125$
$\frac{5}{9}x^2 = 125$
Multiply both sides by $\frac{9}{5}$:
$x^2 = 125 \times \frac{9}{5}$
$x^2 = 25 \times 9$
$x^2 = 225$
Since x is a positive real number, we take the positive square root:
$x = \sqrt{225}$
$x = 15$

Now find the value of y using $y = \frac{2}{3}x$:
$y = \frac{2}{3} \times 15$
$y = 2 \times 5$
$y = 10$

We need to find the value of $xy$:
$xy = 15 \times 10$
$xy = 150$

We should also check if x and y are positive real numbers. $x=15$ and $y=10$ are indeed positive real numbers.

Let’s verify the original equations:
Equation 1: $\log_{5}{(15 + 10)} + \log_{5}{(15 – 10)} = \log_{5}{25} + \log_{5}{5} = 2 + 1 = 3$. This is correct.
Equation 2: $\log_{2}{10} – \log_{2}{15} = \log_{2}{\frac{10}{15}} = \log_{2}{\frac{2}{3}}$.
We are given $1 – \log_{2}{3} = \log_{2}{2} – \log_{2}{3} = \log_{2}{\frac{2}{3}}$. This is also correct.

The value of $xy$ is 150.

Comparing this to the options:
Option A: 150
Option B: 25
Option C: 100
Option D: 250

The calculated value matches Option A.

The final answer is $\boxed{150}$.

Correct_Option:A

Ans: A

Explanation:For the expression $\sqrt{\log_{e}{\dfrac{4x – x^2}{3}}}$ to be a real number, two conditions must be met:
1. The argument of the square root must be non-negative.
2. The argument of the logarithm must be positive.

Condition 1: The argument of the square root is $\log_{e}{\dfrac{4x – x^2}{3}}$. For this to be non-negative, we must have:
$\log_{e}{\dfrac{4x – x^2}{3}} \geq 0$
Since the base of the logarithm is $e > 1$, this inequality is equivalent to:
$\dfrac{4x – x^2}{3} \geq e^0$
$\dfrac{4x – x^2}{3} \geq 1$
Multiply both sides by 3:
$4x – x^2 \geq 3$
Rearrange the terms to form a quadratic inequality:
$x^2 – 4x + 3 \leq 0$
To solve this quadratic inequality, we find the roots of the quadratic equation $x^2 – 4x + 3 = 0$.
Factoring the quadratic, we get $(x-1)(x-3) = 0$.
The roots are $x=1$ and $x=3$.
Since the coefficient of $x^2$ is positive, the parabola opens upwards. The inequality $x^2 – 4x + 3 \leq 0$ holds for values of $x$ between the roots (inclusive).
So, $1 \leq x \leq 3$.

Condition 2: The argument of the logarithm is $\dfrac{4x – x^2}{3}$. For this to be positive, we must have:
$\dfrac{4x – x^2}{3} > 0$
Multiply both sides by 3:
$4x – x^2 > 0$
Factor out $x$:
$x(4 – x) > 0$
To solve this inequality, we find the roots of $x(4-x) = 0$, which are $x=0$ and $x=4$.
The product $x(4-x)$ is positive when $x$ is between the roots (exclusive).
So, $0 < x < 4$.

Now, we need to find the values of $x$ that satisfy both Condition 1 and Condition 2. We need to find the intersection of the intervals $[1, 3]$ and $(0, 4)$.
The intersection of $[1, 3]$ and $(0, 4)$ is $[1, 3]$.

Therefore, the expression is a real number if and only if $1 \leq x \leq 3$.

Correct_Option: A

Ans: D

Provided is the solution to a quantitative aptitude question.
We are given:
$2^{x}=3^{\log_{5}{2}}$
Using the property $a^{\log_{b}{c}} = c^{\log_{b}{a}}$, we can rewrite the right side of the equation:
$3^{\log_{5}{2}} = 2^{\log_{5}{3}}$
So, the equation becomes:
$2^{x}=2^{\log_{5}{3}}$
Since the bases are equal, the exponents must be equal:
$x=\log_{5}{3}$
To express this in a different form, we can use the property of logarithms $\log_{b}(m \times n) = \log_{b}{m} + \log_{b}{n}$. We can rewrite 3 as $\frac{3 \times 5}{5}$:
$x=\log_{5}{\dfrac{3 \times 5}{5}}$
Applying the logarithm property:
$x=\log_{5}{5}+\log_{5}{\dfrac{3}{5}}$
We know that $\log_{b}{b} = 1$:
$x=1+\log_{5}{\dfrac{3}{5}}$
Therefore, the expression for x is $1+\log_{5}{\dfrac{3}{5}}$. This corresponds to option D.

Ans: D

Starting with the given:
$\log_{12}{81}=p$
This implies:
$\log_{81}{12}=\dfrac{1}{p}$
We can rewrite $\log_{81}{12}$ as:
$\log_{3^4}{(3 \times 4)}=\dfrac{4}{p}$
Using logarithm properties, this becomes:
$1+\log_{3}{4}=\dfrac{4}{p}$
Now, applying the Componendo and Dividendo rule to the equation:
$\dfrac{1+\log_{3}{4}-1}{1+\log_{3}{4}+1}=\dfrac{4-p}{4+p}$
Simplifying the left side:
$\dfrac{\log_{3}{4}}{2+\log_{3}{4}}=\dfrac{4-p}{4+p}$
Recognizing that $2 = \log_3{9}$, we can write the denominator as:
$\dfrac{\log_{3}{4}}{\log_{3}{9}+\log_{3}{4}}=\dfrac{4-p}{4+p}$
Using the product rule for logarithms:
$\dfrac{\log_{3}{4}}{\log_{3}{(9 \times 4)}}=\dfrac{4-p}{4+p}$
$\dfrac{\log_{3}{4}}{\log_{3}{36}}=\dfrac{4-p}{4+p}$
Rearranging the expression:
$3 \times \dfrac{4-p}{4+p}=\dfrac{3\log_{3}{4}}{\log_{3}{36}}$
Using the power rule for logarithms:
$3 \times \dfrac{4-p}{4+p}=\dfrac{\log_{3}{(4^3)}}{\log_{3}{36}}$
$3 \times \dfrac{4-p}{4+p}=\dfrac{\log_{3}{64}}{\log_{3}{36}}$
Using the change of base formula for logarithms:
$3 \times \dfrac{4-p}{4+p}=\log_{36}{64}$
We can further simplify $\log_{36}{64}$:
$\log_{36}{64} = \log_{6^2}{8^2}$
Using the property $\log_{a^m}{b^n} = \dfrac{n}{m}\log_a{b}$:
$\log_{6^2}{8^2} = \dfrac{2}{2}\log_{6}{8} = \log_{6}{8}$
Therefore, we have:
$3 \times \dfrac{4-p}{4+p}=\log_{6}{8}$
This leads to the correct answer, option D.

Ans: D

Provided are two equations:
$x^{2018}y^{2017}=\frac{1}{2}$ (Equation 1)
$x^{2016}y^{2019}=8$ (Equation 2)

Divide Equation 2 by Equation 1:
$\dfrac{x^{2016}y^{2019}}{x^{2018}y^{2017}} = \dfrac{8}{1/2}$
$\dfrac{y^{2019-2017}}{x^{2018-2016}} = 16$
$\dfrac{y^2}{x^2} = 16$
Taking the square root of both sides:
$\dfrac{y}{x} = \pm 4$

**Scenario 1: When $\dfrac{y}{x} = 4$**
From this, we have $y = 4x$. Substitute this into Equation 1:
$x^{2018}(4x)^{2017}=\dfrac{1}{2}$
$x^{2018} \cdot 4^{2017} \cdot x^{2017}=\dfrac{1}{2}$
$x^{2018+2017} \cdot 4^{2017}=\dfrac{1}{2}$
$x^{4035} \cdot (2^2)^{2017}=\dfrac{1}{2}$
$x^{4035} \cdot 2^{4034}=\dfrac{1}{2}$
$x^{4035} = \dfrac{1}{2 \cdot 2^{4034}}$
$x^{4035} = \dfrac{1}{2^{4035}}$
Thus, $x = \dfrac{1}{2}$.
Since $y = 4x$, $y = 4 \cdot \dfrac{1}{2} = 2$.
The expression to evaluate is $x^2 + y^3$:
$(\dfrac{1}{2})^2 + (2)^3 = \dfrac{1}{4} + 8 = \dfrac{1+32}{4} = \dfrac{33}{4}$

**Scenario 2: When $\dfrac{y}{x} = -4$**
From this, we have $y = -4x$. Substitute this into Equation 1:
$x^{2018}(-4x)^{2017}=\dfrac{1}{2}$
$x^{2018} \cdot (-4)^{2017} \cdot x^{2017}=\dfrac{1}{2}$
Since 2017 is an odd number, $(-4)^{2017} = -(4^{2017})$.
$x^{2018+2017} \cdot (-4^{2017})=\dfrac{1}{2}$
$x^{4035} \cdot -(2^2)^{2017}=\dfrac{1}{2}$
$x^{4035} \cdot -2^{4034}=\dfrac{1}{2}$
$x^{4035} = \dfrac{1}{2 \cdot (-2^{4034})}$
$x^{4035} = \dfrac{-1}{2^{4035}}$
Thus, $x = \dfrac{-1}{2}$.
Since $y = -4x$, $y = -4 \cdot (\dfrac{-1}{2}) = 2$.
The expression to evaluate is $x^2 + y^3$:
$(\dfrac{-1}{2})^2 + (2)^3 = \dfrac{1}{4} + 8 = \dfrac{1+32}{4} = \dfrac{33}{4}$

In both scenarios, the value of $x^2 + y^3$ is $\dfrac{33}{4}$.

Ans: A

Explanation:We are given two equations:
1. $\log_{2}({5+\log_{3}{a}})=3$
2. $\log_{5}({4a+12+\log_{2}{b}})=3$

Let’s solve the first equation for ‘a’.
From the definition of logarithm, if $\log_x{y} = z$, then $x^z = y$.
Applying this to the first equation:
$2^3 = 5+\log_{3}{a}$
$8 = 5+\log_{3}{a}$
Subtract 5 from both sides:
$8 – 5 = \log_{3}{a}$
$3 = \log_{3}{a}$
Now, apply the definition of logarithm again:
$3^3 = a$
$a = 27$

Now let’s use the value of ‘a’ to solve the second equation for ‘b’.
Substitute $a=27$ into the second equation:
$\log_{5}({4(27)+12+\log_{2}{b}})=3$
$\log_{5}({108+12+\log_{2}{b}})=3$
$\log_{5}({120+\log_{2}{b}})=3$
Apply the definition of logarithm:
$5^3 = 120+\log_{2}{b}$
$125 = 120+\log_{2}{b}$
Subtract 120 from both sides:
$125 – 120 = \log_{2}{b}$
$5 = \log_{2}{b}$
Apply the definition of logarithm again:
$2^5 = b$
$b = 32$

We need to find the value of $a+b$.
$a+b = 27+32$
$a+b = 59$

Comparing this result with the given options:
Option A: 59
Option B: 40
Option C: 32
Option D: 67

The calculated value of $a+b$ matches Option A.

The final answer is $\boxed{59}$.
Correct_Option:A

Ans: 5

For $x = 0$, $2^x$ equals 1, which falls within the specified interval of [0.25, 200]. Evaluating $2^x + 2$ gives $1 + 2 = 3$. Since 3 is divisible by 3, $x = 0$ is a valid possibility.

For $x = 1$, $2^x$ equals 2, which is within the interval [0.25, 200]. Calculating $2^x + 2$ results in $2 + 2 = 4$. As 4 is divisible by 4, $x = 1$ is a valid possibility.

For $x = 2$, $2^x$ equals 4, which is within the interval [0.25, 200]. Evaluating $2^x + 2$ gives $4 + 2 = 6$. Since 6 is divisible by 3, $x = 2$ is a valid possibility.

For $x = 3$, $2^x$ equals 8, which is within the interval [0.25, 200]. Calculating $2^x + 2$ results in $8 + 2 = 10$. As 10 is not divisible by 3 or 4, $x = 3$ is not a valid solution.

For $x = 4$, $2^x$ equals 16, which is within the interval [0.25, 200]. Evaluating $2^x + 2$ gives $16 + 2 = 18$. Since 18 is divisible by 3, $x = 4$ is a valid possibility.

For $x = 5$, $2^x$ equals 32, which is within the interval [0.25, 200]. Calculating $2^x + 2$ results in $32 + 2 = 34$. As 34 is not divisible by 3 or 4, $x = 5$ is not a valid solution.

For $x = 6$, $2^x$ equals 64, which is within the interval [0.25, 200]. Evaluating $2^x + 2$ gives $64 + 2 = 66$. Since 66 is divisible by 3, $x = 6$ is a valid possibility.

For $x = 7$, $2^x$ equals 128, which is within the interval [0.25, 200]. Calculating $2^x + 2$ results in $128 + 2 = 130$. As 130 is not divisible by 3 or 4, $x = 7$ is not a valid solution.

For $x = 8$, $2^x$ equals 256, which is outside the interval [0.25, 200]. Therefore, $x$ cannot be 8 or any value greater than 7.

The set of all possible integer values for $x$ is {0, 1, 2, 4, 6}. This means $x$ can take on 5 distinct integer values.

Ans: D

Given the inequality:
$4^{n} > 17^{19}$
We can rewrite the left side of the inequality:
$16^{n/2} > 17^{19}$
Since $16$ is less than $17$, for the inequality to hold true, the exponent on the left side must be greater than the exponent on the right side. Therefore, we can deduce:
$n/2 > 19$
Multiplying both sides by 2, we get:
$n > 38$
Consequently, the option that satisfies this condition is the correct choice.

Ans: A

Explanation:We are asked to find the value of the given expression:
$ \frac{1}{log_{2}100}-\frac{1}{log_{4}100}+\frac{1}{log_{5}100}-\frac{1}{log_{10}100}+\frac{1}{log_{20}100}-\frac{1}{log_{25}100}+\frac{1}{log_{50}100} $
We can use the change of base formula for logarithms, which states that $\frac{1}{log_{b}a} = log_{a}b$.
Applying this formula to each term in the expression, we get:
$ log_{100}2 – log_{100}4 + log_{100}5 – log_{100}10 + log_{100}20 – log_{100}25 + log_{100}50 $
Now we can use the properties of logarithms: $log_{a}b – log_{a}c = log_{a}\left(\frac{b}{c}\right)$ and $log_{a}b + log_{a}c = log_{a}(b \times c)$.
We can group the positive and negative terms:
$ (log_{100}2 + log_{100}5 + log_{100}20 + log_{100}50) – (log_{100}4 + log_{100}10 + log_{100}25) $
Combine the terms within each parenthesis:
$ log_{100}(2 \times 5 \times 20 \times 50) – log_{100}(4 \times 10 \times 25) $
Calculate the products inside the logarithms:
$ 2 \times 5 \times 20 \times 50 = 10 \times 1000 = 10000 $
$ 4 \times 10 \times 25 = 40 \times 25 = 1000 $
So the expression becomes:
$ log_{100}10000 – log_{100}1000 $
Now we can evaluate these logarithms. We need to find the power to which 100 must be raised to get 10000, and the power to which 100 must be raised to get 1000.
$100^x = 10000 \implies (10^2)^x = 10^4 \implies 10^{2x} = 10^4 \implies 2x = 4 \implies x = 2$.
So, $log_{100}10000 = 2$.
$100^y = 1000 \implies (10^2)^y = 10^3 \implies 10^{2y} = 10^3 \implies 2y = 3 \implies y = \frac{3}{2}$.
So, $log_{100}1000 = \frac{3}{2}$.
The expression is:
$ 2 – \frac{3}{2} = \frac{4}{2} – \frac{3}{2} = \frac{1}{2} $

Alternatively, we can directly apply the logarithm properties without evaluating individual logarithms:
$ log_{100}2 – log_{100}4 + log_{100}5 – log_{100}10 + log_{100}20 – log_{100}25 + log_{100}50 $
$ = log_{100}\left(\frac{2 \times 5 \times 20 \times 50}{4 \times 10 \times 25}\right) $
$ = log_{100}\left(\frac{10000}{1000}\right) $
$ = log_{100}(10) $
Since $100^{1/2} = \sqrt{100} = 10$, we have $log_{100}(10) = \frac{1}{2}$.

The final answer is $\frac{1}{2}$.

Correct_Option:A

Ans: A

Explanation:Given the relation $p^3 = q^4 = r^5 = s^6$.
Let’s set this common value to a constant, say k.
$p^3 = k \implies p = k^{1/3}$
$q^4 = k \implies q = k^{1/4}$
$r^5 = k \implies r = k^{1/5}$
$s^6 = k \implies s = k^{1/6}$

We need to find the value of $log_{s}{(pqr)}$.
First, let’s find the product $pqr$:
$pqr = k^{1/3} \cdot k^{1/4} \cdot k^{1/5}$
When multiplying powers with the same base, we add the exponents:
$pqr = k^{\frac{1}{3} + \frac{1}{4} + \frac{1}{5}}$
To add the fractions, we find a common denominator, which is 60:
$\frac{1}{3} + \frac{1}{4} + \frac{1}{5} = \frac{20}{60} + \frac{15}{60} + \frac{12}{60} = \frac{20+15+12}{60} = \frac{47}{60}$
So, $pqr = k^{\frac{47}{60}}$.

Now we need to find $log_{s}{(pqr)}$.
We know that $s = k^{1/6}$.
We can express $pqr$ in terms of s.
From $s^6 = k$, we have $k = s^6$.
Substitute this value of k into the expression for pqr:
$pqr = (s^6)^{\frac{47}{60}} = s^{6 \cdot \frac{47}{60}} = s^{\frac{47}{10}}$

Now, we can find $log_{s}{(pqr)}$:
$log_{s}{(pqr)} = log_{s}{(s^{\frac{47}{10}})}$
Using the logarithm property $log_b{(b^x)} = x$:
$log_{s}{(s^{\frac{47}{10}})} = \frac{47}{10}$

Let’s check the options.
Option A: $\frac{47}{10}$

The final answer is $\frac{47}{10}$.

Correct_Option:A

Ans: D

Given that $\log_3 x = a$ and $\log_{12} y = a$.
This implies that $x = 3^a$ and $y = 12^a$.
The geometric mean of $x$ and $y$ is defined as $\sqrt{x \times y}$.
Substituting the values of $x$ and $y$, we get $\sqrt{3^a \times 12^a}$.
This expression simplifies to $(3 \times 12)^a$ raised to the power of $1/2$, which is $(36)^a$ raised to the power of $1/2$, resulting in $(6^2)^a$ raised to the power of $1/2$, which equals $6^a$.
Therefore, the geometric mean, denoted by $G$, is $6^a$.
From $G = 6^a$, we can rewrite this in logarithmic form as $\log_6 G = a$.

Ans: C

Explanation:We need to evaluate the expression $\log_{0.008}\sqrt{5}+\log_{\sqrt{3}}81-7$.

First, let’s simplify each term separately.

Term 1: $\log_{0.008}\sqrt{5}$
We can write $0.008$ as a power of 10: $0.008 = \frac{8}{1000} = \frac{1}{125} = \frac{1}{5^3} = 5^{-3}$.
We can write $\sqrt{5}$ as $5^{1/2}$.
So, $\log_{0.008}\sqrt{5} = \log_{5^{-3}}5^{1/2}$.
Using the property $\log_{a^m}b^n = \frac{n}{m}\log_a b$, we get:
$\log_{5^{-3}}5^{1/2} = \frac{1/2}{-3}\log_5 5 = \frac{1/2}{-3} \times 1 = -\frac{1}{6}$.

Term 2: $\log_{\sqrt{3}}81$
We can write $\sqrt{3}$ as $3^{1/2}$.
We can write $81$ as $3^4$.
So, $\log_{\sqrt{3}}81 = \log_{3^{1/2}}3^4$.
Using the property $\log_{a^m}b^n = \frac{n}{m}\log_a b$, we get:
$\log_{3^{1/2}}3^4 = \frac{4}{1/2}\log_3 3 = \frac{4}{1/2} \times 1 = 8$.

Now, substitute these values back into the original expression:
$\log_{0.008}\sqrt{5}+\log_{\sqrt{3}}81-7 = -\frac{1}{6} + 8 – 7$.
$-\frac{1}{6} + 8 – 7 = -\frac{1}{6} + 1$.
To add these, find a common denominator:
$-\frac{1}{6} + \frac{6}{6} = \frac{-1+6}{6} = \frac{5}{6}$.

Therefore, the value of the expression is $\frac{5}{6}$.

Comparing this with the given options:
Option A: 1/3
Option B: 2/3
Option C: 5/6
Option D: 7/6

The calculated value matches Option C.

Correct_Option:C

Ans: D

Explanation:We are given the equation $\log_{3}5 = \log_{5}(2 + x)$.
Let $k = \log_{3}5$.
Then, by the definition of logarithm, $3^k = 5$.
The given equation can be rewritten as $k = \log_{5}(2 + x)$.
By the definition of logarithm, $5^k = 2 + x$.
We know that $k = \log_{3}5$. Substitute this into the equation $5^k = 2 + x$:
$5^{\log_{3}5} = 2 + x$.

Let’s analyze the value of $k = \log_{3}5$.
Since $3^1 = 3$ and $3^2 = 9$, we know that $1 < \log_{3}5 < 2$.
So, $1 < k < 2$.

Now let’s analyze $5^k$.
Since $1 < k < 2$, we can say that $5^1 < 5^k < 5^2$.
This means $5 < 5^k < 25$.

We have $5^k = 2 + x$.
So, $5 < 2 + x < 25$.
Subtracting 2 from all parts of the inequality, we get:
$5 – 2 < x < 25 - 2$
$3 < x < 23$.

Now let’s check the given options:
Option A: 0 < x < 3
Option B: 23 < x < 30
Option C: x > 30
Option D: 3 < x < 23

Our derived range for x is $3 < x < 23$, which exactly matches Option D.

To be more precise about the bounds:
We know $\log_3 3 = 1$ and $\log_3 9 = 2$. Since $3 < 5 < 9$, we have $1 < \log_3 5 < 2$.
Let $k = \log_3 5$. So $1 < k < 2$.
The equation is $k = \log_5 (2+x)$.
This means $5^k = 2+x$.
Since $1 < k < 2$, we have $5^1 < 5^k < 5^2$.
So, $5 < 5^k < 25$.
Substituting $5^k = 2+x$, we get $5 < 2+x < 25$.
Subtracting 2 from all parts, we get $5-2 < x < 25-2$.
Therefore, $3 < x < 23$.

Correct_Option:D

Ans: A

The problem states that $9^{x-\frac{1}{2}}-2^{2x-2}=4^{x}-3^{2x-3}$.
We can simplify the equation by expressing all terms as powers of $3$ and $2$.
This transforms the original equation into $3^{2x-1} + 3^{2x-3} = 2^{2x} + 2^{2x-2}$.
By factoring, we get $3^{2x-3} \times 10 = 2^{2x-2} \times 5$.
Dividing both sides by $5$, we arrive at $3^{2x-3} = 2^{2x-3}$.
For this equality to hold, the exponents must be equal to zero, leading to $2x-3=0$, which means $x=3/2$.

Ans: 3

$log(2^{a}\times3^{b}\times5^{c} )$ = $ \frac{log ( 2^{2}\times3^{3}\times5^{1}) + log(2^{6}\times3^{1}\times5^{7} ) + log(2^{1}\times3^{2}\times5^{4} ) }{3} $
$log(2^{a}\times3^{b}\times5^{c} )$ = $ \frac{log ( 2^{2+6+1}\times3^{3+1+2}\times5^{1+7+4}) }{3} $
$log(2^{a}\times3^{b}\times5^{c} )$ = $ \frac{log ( 2^{9}\times3^{6}\times5^{12}) }{3} $
$3log(2^{a}\times3^{b}\times5^{c} )$ = $ log ( 2^{9}\times3^{6}\times5^{12}) $
Thus, 3a = 9, which implies a = 3