Is this logarithm-based way to solve an exponential equation valid? A friend of mine solved the following exponential equation in the following way:
$$4^{x+1}-8^{2x}=0$$
Since $\log_{4}8=\frac{3}{2}$, $4^{x+1}-(4^{2x})^{\frac{3}{2}})=0 \Leftrightarrow 4^{x+1}-4^{3x}=0$. Now that both $x$-terms are base 4, this equation is trivial to solve. However, I don't understand how that process can work. Is there any base to this process or was it just a lucky coincidence?
Note: I know how to solve that equation, I'm only asking if my friend's approach is valid or not

Well you have that $8=2^{3}=(\sqrt{4})^3=4^{3/2}$.Now as to why it works we have that
$$a^{\log_a b}=b$$
Now plugging in $b=8$ we can write
$$8=4^{\log_4 8}$$
And we have that $\log_4 8=\frac{3}{2}$
So we can write $8^{2x}$ as $(4^{3/2})^{2x}$.
Your friends approach is correct.
