can somebody prove $ a^{\log(a)^x} = x$

Question

can somebody please prove that $$ a^{\log(a)^x} = x$$

it is written in my text book but I cant seem to get it

Some examples if possible

Answer 1

Start with the definition of $b = \log_{a}x \implies x=a^{b} \implies \ln x = b\ln a$

Then let $c = a^b =a^{\log_{a}x}$

Then $\ln c = \ln(a^{b}) = b\ln a$

Note that $\ln x = b\ln a$. Then by comparison, it follows that $c=x = a^{\log_{a}x}$ as required

Or you could think of it in words. $\log_{a}x$ means "the power of $a$ such that $a$ raised to that power equals $x$".

Then $a^{\log_{a}x}$ can be thought of as: "$a$ to the power of (the power of $a$ that gives $x$)" which must then give $x$

Answer 2

It rather depends on what properties of logarithms you wish to use.

For example, with $a \gt 0$ and $a \not = 1$,

$f(y)=a^y$ and $g(x)=\log_a x$ are inverse bijective functions, with $f: \mathbb{R}\to \mathbb{R}^+$ and $g: \mathbb{R}^+\to \mathbb{R}$

so $f(g(x))=x$, which means $a^{\log_a x} = x$

---

Alternatively, using natural logarithms,

$\log_a x= \frac{\log_e x}{\log_e a}$ and $a^y= e^{y \log_e a}$

so $a^{\log_a x}= e^{(\log_a x)(\log_e a)} = e^{\log_e x} = x$