Question:
$$ (3+4)(3^2+4^2)(3^4+4^4)(3^8+4^8)(3^{16}+4^{16})(3^{32}+4^{32}) = 4^x - 3^x $$
Find $x$.
Proposition:
For $$ p = a^x - b^x $$ We assume $x = \log_r p$, then.
$$ p = a^{\log_r p} - b^{\log_r p} \\ p = p^{\log_r a} - p^{\log_r b}$$ Taking log both sides yield. $$1 = \log_r a/b \Rightarrow r = \frac ab$$
So, using the proposition for $p= LHS$ of the question, we get the answer, $$x = \log_{4/3} p$$ Which is certainly not equal to the answer, which is 64. So, what is the problem with logarithm?
Edit - For $$ p = a^{\log_{r} p} - b^{\log_{r} p} $$ There can exist an unique $r$ for which the equation can statisfy the equation. There shouldn't be a problem with it.
The equation $x = \log_{a/b} p$ is equivalent to $p = (a/b)^x = \dfrac{a^x}{b^x}$, so you're suggesting that $a^x-b^x = \dfrac{a^x}{b^x}$, which is certainly not the case in general. Your error was in the step 'taking log both sides yield'—you applied the laws of logarithms incorrectly.
Instead of trying to 'solve for $x$', a better way to approach this problem would be to multiply the left-hand side by $1$, but write $1$ as $4-3$. Then repeatedly apply the 'difference of two squares' rule. For example $$ (4-3)(3+4) = (4-3)(4+3) = 4^2-3^2 $$ and so $$ (4-3)(3+4)(3^2+4^2) = (4^2-3^2)(4^2+3^2) = 4^4-3^4$$ and so on.