How do I solve for n?
$125 = x * 2^n$
This is what I have so far:
$5^3 = x * 2^n$
I do remember that according to the exponential rules, that the powers should be the same if the equation is like this:
$8 = 2^n$
$2^3 = 2^n \iff 2^3 = 2^3$
I am not sure if this rule can be used in the equation above.
On
You cannot solve for $n$, unless you want your answer to be in terms of $x$. In that case, $125/x=2^n$ tells you $n=\log_2(125/x)=3\log_2(5)-\log_2(x)$.
The reason it's not possible to solve for $n$ is because, no matter what $n$ you choose, there's always an appropriate choice of $x$ so that the equation is satisfied. (Why? Can you find an expression for such an $x$?) On the other hand, if $x$ is restricted to an integer, then you want to consider prime factorizations of both sides. I'll leave this to you.
Hint: We get $$\frac{125}{x}=2^n$$ so $$\ln\left(\frac{125}{x}\right)=n\ln(2)$$