TITLE: How to solve equations like $8^{2n+1} = 32^{n+1}$
QUESTION [6 upvotes]: I have stumbled on this question, and there are a few questions after it of the same type. How do I solve it and what is the right approach for this kind of question?
$$8^{2n+1} = 32^{n+1}$$
I need to find the value of $^n$. Step by step if at all possible please. My approach was to start like:
$$(2^3)^{2n+1} = (2^3 \times 2^2)^{n+1}$$
Then I was lost as to how to proceed. Do I need some sort of prerequisite knowledge that I am missing before attempting this? The book has only shown $7$ laws of indices before asking this question.

REPLY [4 votes]: The other solutions to this problem presuppose two factors:
1) the two bases are themselves both powers of some common base;
2) the solver recognizes that fact.
But suppose that the bases 8 and 32 in the problem above were replaced with 2197 and 371293.  Or even worse, with 2197 and 317293.
A more "bullet-proof" technique would be to get the unknown out of the exponents by taking logarithms of both sides.  The equation to be solved becomes:$$(2n+1)\log(8)=(n+1)\log (32)$$When you divide log32 by log8 and get $\frac{5}{3}$, it's clear that you missed some subtle point. but you're already well on your way to the solution, and the next time the numbers might not work out so neatly...