# Do you know this property of the logarithm?

1. Apr 29, 2012

### azabak

Playing around with logarithms I found an interesting property that "log b^n(a^n) = log b(a)". Then I tried to find some kind of proof that this is right and not only a coincidence. Ι made a gereral formula for any value of both n's (α and β) so that "log b^β(a^α) = x". Therefore "a^α = b^(β*x)" ; "a = b^(β*x/α)" ; "log b(a) = β*x/α" ; "x = (α/β)*log b(a)". And therefore "log b^β(a^α) = (α/β)*log b(a)".

2. Apr 29, 2012

### Stephen Tashi

It's unclear what that notation is supposed to mean.

Can you write out what "log b^n(a^n)" means in words? Or perhaps master the forums LaTex: https://www.physicsforums.com/showthread.php?t=546968

3. Apr 29, 2012

### HallsofIvy

I think you mean "$log_{b^n}(a^n)= log_b(a)$". That is, that the logarithm, base $b^n$, of $a^n$ is the same as the logarithm, base b, of a. (Of course, a and b must be positive.)

If $y= log_{b^n}(a^n)$ then $a^n= (b^n)^y= b^{ny}= (b^y)^n$. Can you complete it now?