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.

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- Feb 6th 2010, 05:48 PMsiouxDifferent 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. - Feb 6th 2010, 05:56 PMfelper
- Feb 6th 2010, 06:03 PMsioux
Haha, was right in front of me. Thanks!

- Feb 7th 2010, 12:27 AMHallsofIvy
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 $\displaystyle c= log_a(b)$.