# Different base logarithm proof

• Feb 6th 2010, 06:48 PM
sioux
Different base logarithm proof
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.
• Feb 6th 2010, 06:56 PM
felper
Quote:

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.

$\log_a(x)=\dfrac{\log(x)}{\log(a)}=\underbrace{\df rac{\log(b)}{\log(a)}}_{c}\cdot \dfrac{\log(x)}{\log(b)}=c\cdot \log_b(x)$
• Feb 6th 2010, 07:03 PM
sioux
Haha, was right in front of me. Thanks!
• Feb 7th 2010, 01:27 AM
HallsofIvy
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 $c= log_a(b)$.