logarithm proof for $a^{log_a(b)}=b$ [duplicate]

Question

I have tried proving for $a^{log_a(b)}=b$ , but I feel is incorrect, so how can I prove this?

I have proved it as follows:

$log_aa^{log_a(b)}=log_ab$

$log_a(b)log_aa= log_ab$

$log_a(b)= log_ab$

Answer 1

It is common to define $\log_a$ as the inverse of the function $a\mapsto a^x$. If we take this approach, then $a^{\log_a(x)}=x$ is part of the definition of what $\log_a$ means, and so it is not appropriate to try to prove this statement.

To understand why, consider that for a function $f$ with a domain of $X$ and range of $Y$, we define its inverse $f^{-1}$ as the unique function with domain $Y$ satisfying $$ f^{-1}(f(x))=x $$ for all $x\in X$, and $$ f(f^{-1}(x))=x $$ for all $x\in Y$. In this case $f=a\mapsto a^x$, $f^{-1}=\log_a$, $X=\Bbb{R}$, and $Y=\Bbb{R^+}$. Therefore, $$ \log_a(a^x)=x $$ for all $x\in\Bbb{R}$, and $$ a^{\log_a(x)}=x $$ for all $x\in\Bbb{R^+}$. Notice also that $f^{-1}(x)$ can be understood to be the answer to the question "what is the unique number $t$ such that $f(t)=x$?". Therefore, $\log_a(x)$ is the answer to the question "what is the unique number $t$ such that $a^t=x$?". Hence, $a^{\log_a(x)}=x$ because that's what $\log_a(x)$ means.