Proof that $b^{\log_b(x)} = x$

Question

I understand that the exponential functions are inverses, and would therefore map $x$ when formed as a composition, but I cannot find any formal mathmatical proofs. My thought process is: $$\log_b(b^{\log_b(x)}) = \log_b(x) \rightarrow \log_b(x)=\log_b(x) \rightarrow x=x$$ Is that the only way of going about it or are there other formal proofs?

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score 4 · Accepted Answer · answered Apr 12 '14 at 14:55

4

This is a tautology since $\log_b(x)$ is defined as the exponent $m$ such that $b^m = x$.

answered Apr 12 '14 at 14:55

user141421

2,538

score 3 · Answer 2 · answered Apr 12 '14 at 14:39

3

By definition, if $A^x=B$, then $x=\log_AB$. Now, replace the value of x from the latter identity into the former.

answered Apr 12 '14 at 14:39

Lucian

48,334
2
83
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Isn't that what I did already? – OpieDopee Apr 12 '14 at 15:08

score 2 · Answer 3 · answered Apr 12 '14 at 14:52

2

What exponent $a$ can you put on $b$ so that $b^a=x$? $\log_b(x)$ is just the name for an $a$ that makes $b^a$ be equal to $x$. So by definition $b^{\log_b(x)}=x$.

answered Apr 12 '14 at 14:52

john.abraham

809

Proof that $b^{\log_b(x)} = x$

3 Answers3

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