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

Another, less convoluted way is to convert everything to base 2:
$$
0 = (2^2)^{x+1} - (2^3)^{2x} = 2^{2x+2} - 2^{6x}...
$$
