Question

The solution set of...
options
a {5}
b {625}
c {25}
d {3125}​

Answer 1

$\large\text{\underline{Formula}}$

The power in the given question can be regarded as,

$\begin{aligned}a^{\log_{b}c}&=a^{\frac{\log_{a}c}{\log_{a}b}}\\&=(a^{\log_{a}c})^{\frac{1}{\log_{a}b}}\\&=(a^{\log_{a}c})^{\log_{b}a}\\&=c^{\log_{b}a}}.\end{aligned}$

We get,

$\cdots\longrightarrow a^{\log_{b}c}=c^{\log_{b}a}.$

$\large\text{\underline{Solution}}$

Now to solve the problem, we apply the formula on one power.

We know that,

$\cdots\longrightarrow2^{\log_{5}x}=x^{\log_{5}{2}}.$

Now, let us solve the exponential equation.

$\cdots\longrightarrow 3x^{\log_{5}2}-x^{\log_{5}{2}}=64$

$\cdots\longrightarrow 2x^{\log_{5}2}=64$

$\cdots\longrightarrow x^{\log_{5}2}=32$

If we take the $\log$ of base 5,

$\cdots\longrightarrow \log_{5}(x^{\log_{5}2})=\log_{5}(2^{5})$

$\cdots\longrightarrow\log_{5}2\log_{5}x=5\log_{5}2$

$\cdots\longrightarrow \log_{5} x=5.$

We know that,

$\cdots\longrightarrow x=5^{5}=3125.$

$\large\text{\underline{Answer}}$

Choice D. {3125} is the correct answer.

$\large\text{\underline{Extra! Question}}$

Simplify $\large a^{\frac{\log(\log a)}{\log a}}(a>1)$.

(Formula)

Consider an equation, $\log_{a}b=x$.

Then, $a^{x}=b$.

If we take $\log$ of base c$(c>0,c\neq1)$ on both sides,

$\cdots\longrightarrow\log_{c}a^{x}=\log_{c}b$

$\cdots\longrightarrow x\log_{c}a=\log_{c}b$

$\cdots\longrightarrow x=\dfrac{\log_{c}b}{\log_{c}a}.$

We get,

$\cdots\longrightarrow \log_{a}b=\dfrac{\log_{c}b}{\log_{c}a}.$

(Solution)

We know that,

$\cdots\longrightarrow\dfrac{\log (\log a)}{\log a}=\log_{a}(\log a).$

So,

$\cdots\longrightarrow a^{\log_{a}(\log a)}=\log a.$