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

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:

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

$$log_a(b)log_aa= log_ab$$

$$log_a(b)= log_ab$$

• Which definition of $\log_a$ do you use? Jun 5, 2021 at 9:35
• 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... Jun 5, 2021 at 9:36
• Can I say the proof I have given is consider ok?
– Joe
Jun 5, 2021 at 9:46
• @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.
– Joe
Jun 5, 2021 at 9:59

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.