Unusual Logarithm Problem I encountered this problem a while ago.
Solve for $n$: $(\log_bn)^m = \log_an$. 
I had two questions about this problem.
1) Is there a real application to the solution to this equation in advanced mathematics or physics?
2) I found a formula that finds all $n$'s except for 1. How could I fix that? 
Here is the formula I found: $n = b^{(\log_ab)^\frac{1}{m-1}}$
 A: 1) The point is to gain skill
in the manipulation of
these kind of expressions.
2) Clearly $n=1$ is a solution.
So I would write
"$n=1$ is a solution.
For the other solutions,
assume $n \ne 1$"
and then proceed with
your solution.
For
$(\log_bn)^m = \log_an$,
I would do this:
Writing in terms of
$\ln$,
and assuming
$n \ne 1$,
this becomes
$(\frac{\ln n}{\ln b})^m = \frac{\ln n}{\ln a}$
or
(this is where the
assumption that
$n \ne 1$
is used)
$(\ln n)^{m-1}
=\frac{(\ln b)^m}{\ln a}
$
or
$\ln n
=(\frac{(\ln b)^m}{\ln a})^{1/(m-1)}
=\ln b(\frac{\ln b}{\ln a})^{1/(m-1)}
=\ln b(\log_a b)^{1/(m-1)}
$
so that
$n
=e^{\ln b(\log_a b)^{1/(m-1)}}
=b^{(\log_a b)^{1/(m-1)}}
$
which confirms your answer.
A: $log_bn=\frac  {log_an}{log_ab} $ by the change of base formula...   so $(\frac {log_an}{log_ab})^m=log_an \implies (log_an)^{m-1}=(log_ab)^m \implies log_an=(log_ab)^{\frac m {m-1}}\implies n=a^{(log_ab)^{\frac m {m-1}}}$ if $m\not=1$...  
If $m=1$, then $log_an=log_bn $, which is true for  $n=1$...
