Different base logarithm proof

**sioux** · February 6th 2010, 05:48 PM

Hello!

I need to prove that $log_a(x) = c * log_b(x)$

I don't seem to know how to start proving this very evident example.

I started to move them around, and change bases, but got nothing useful.

**felper** · February 6th 2010, 05:56 PM

Originally Posted by sioux

Hello!

I need to prove that $log_a(x) = c * log_b(x)$

I don't seem to know how to start proving this very evident example.

I started to move them around, and change bases, but got nothing useful.

$\log_a(x)=\dfrac{\log(x)}{\log(a)}=\underbrace{\df rac{\log(b)}{\log(a)}}_{c}\cdot \dfrac{\log(x)}{\log(b)}=c\cdot \log_b(x)$

**sioux** · February 6th 2010, 06:03 PM

Haha, was right in front of me. Thanks!

**HallsofIvy** · February 7th 2010, 12:27 AM

If you did not know that formula, you could use the fact that $log_a(x)$ is defined as the inverse function to $a^x$ . That is, that $log_a(a^x)= x$ and $a^{log_a(x)}= x$ .

In particular, if $y= log_b(x)$ then $x= b^y$ . But then $x= a^{log_a(b^y)}= a^{ylog_a()}$ . From that, $log_a(x)= log_a(a^{ylog_a(b)}= (log_a(b))y= (log_a(b))log_b(x)$ . That is $log_a(x)= c log_b(x)$ with $c= log_a(b)$ .

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