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: $n = b^{(\log_ab)^\frac{1}{m-1}}$
On
$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 $n=1$...
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.