# Thread: Different base logarithm proof

1. ## Different base logarithm proof

Hello!

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

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.

2. Originally Posted by sioux
Hello!

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

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.

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

3. Haha, was right in front of me. Thanks!

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

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