Simple logarithms #1

Suppose $\large x = a^{\sqrt{\log_{a}(b)}}.$ Then by the change of base rule

$\large \log_{a}(x) = \sqrt{\log_{a}(b)} \Longrightarrow \dfrac{\log_{b}(x)}{\log_{b}(a)} = \dfrac{1}{\sqrt{\log_{b}(a)}}$

$\large \Longrightarrow \log_{b}(x) = \sqrt{\log_{b}(a)} \Longrightarrow x = b^{\sqrt{\log_{b}(a)}}.$

The given expression is then equal to $\large x - x = \boxed{0}.$

First we will focus on $\large{a}^{\sqrt{\log_{a}{b}}}$ , which can be expessed as $\large{a}^{\sqrt{\frac{\ln{b}}{\ln{a}}}}$ according to the change of base rule.

Rationalizing the denominator, we get:

$\large{a}^{\frac{\sqrt{\ln{b}}}{\sqrt{\ln{a}}}} =\large{a}^{\frac{\sqrt{\ln{b}\ln{a}}}{\ln{a}}}$ which can also be expresed as $\large(\large{a}^{\frac{1}{\ln{a}}})^{\sqrt{\ln{a}\ln{b}}}$

Now, by the change of base rule, we have:

$\ln{a}=\large{\frac{\log_{a}{a}}{\log_{a}{e}}}=\large{\frac{1}{\log_{a}{e}}} \Rightarrow \frac{1}{\ln{a}}=\log_{a}{e}$

Hence $(\large{a}^{\frac{1}{\ln{a}}})^{\sqrt{\ln{a}\ln{b}}}=(\large{a}^{\log_{a}{e}})^{\sqrt{\ln{a}\ln{b}}}= \large{e}^{\sqrt{\ln{a}\ln{b}}}$

A similar simplification holds for $\large{b}^{\sqrt{\log_{b}{a}}}$ , so the original expresion reduces to

$\large{e}^{\sqrt{\ln{a}\ln{b}}}-\large{e}^{\sqrt{\ln{a}\ln{b}}}=\boxed{0}$