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I'm very new to math, I'm sorry if my question is stupid. I started to study math by my own so I can study Computer Engeneering.

I'm studying logarithms and I try to come up with simple proofs of the properties I learn as often as possible. In trying to make sense of the change of base formula, I came up with the following reasoning:

The change of base formula is $\log_ab=\log_cb/\log_ca$. Assuming I can always write a number $a, a>0$ as any number $c, c>0$ to the power of some number ($a=c^x$):

$$y=\log_ab=\frac{\log_c((c^x)^y)}{\log_c(c^x)}=\frac{xy}x=y$$

I've seen other -and better- proofs, but that is the one I thought. The question is: how can I be sure that my assumption is true?

Sorry for any mistakes, English is not my native language.

Rrmm
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2 Answers2

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Hint
What happens if you chose $a=0$ and $c\ne 0$?

Edited Question
With some further restrictions, this is indeed true. If both $a$ and $c$ are positive real numbers then there exists a unique real number $x$ such that $$a=c^x$$ This number is defined as $$x = \log_c a$$ This also means that any proof of this fact cannot use the logarithm since that would create a tautology.

AlexR
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  • In that case it won't work. I rephrased the question, thanks. – Rrmm Feb 01 '15 at 02:05
  • @Rrmm You'll also want to demand $c>0$ for that matter. – AlexR Feb 01 '15 at 02:23
  • In that case I would like to know how can you prove that given the restrictions mentioned $log_c a$ is always a real number. – Rrmm Feb 01 '15 at 02:34
  • @Rrmm That question is basically an entirely new one. I suggest you ask a follow-up question because changing the meaning of an answered question is frowned upon for obvious reasons. – AlexR Feb 01 '15 at 02:35
  • I'll do that. Thanks for your time. – Rrmm Feb 01 '15 at 02:39
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If $c>1$, then the function $x\mapsto c^x$ is continuous, so it takes all values between its limit in $-\infty$, which is $0$, and its limit in $+\infty$, which is $+\infty$.

If $0<c<1$ then $c^x=\left(\displaystyle\frac1c\right)^{-x}$, so again $c^x$ takes all positive real values.

Berci
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