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:


$log_a(b)log_aa= log_ab$

$log_a(b)= log_ab$

  • 3
    $\begingroup$ Which definition of $\log_a$ do you use? $\endgroup$
    – Paul Frost
    Jun 5, 2021 at 9:35
  • 4
    $\begingroup$ By definition, $\; \log_ax\;$ is the power to which the basis $\;a\;$ must be raised in order to get $\;x\;$ . With this, $\;a^{\log_ab}=b\;$ is completely trivial... $\endgroup$
    – DonAntonio
    Jun 5, 2021 at 9:36
  • $\begingroup$ Can I say the proof I have given is consider ok? $\endgroup$
    – Joe
    Jun 5, 2021 at 9:46
  • 1
    $\begingroup$ @Joe: It doesn't make sense to try to prove this statement. This would be akin to trying to prove that $\pi$ is the ratio of a circle's circumference to its diameter. That is the definition of what $\pi$ actually means. $\endgroup$
    – Joe
    Jun 5, 2021 at 9:59

1 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.

  • $\begingroup$ @user21820: Okay, I've rewritten my answer. What do you think? $\endgroup$
    – Joe
    Aug 1, 2021 at 13:40
  • $\begingroup$ I fixed a small error, and it's now great, so I'll remove my now irrelevant comment! =) $\endgroup$
    – user21820
    Aug 1, 2021 at 14:16

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