$$\log_3(5(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})+2^{64})=?$$
At a first glance, it seems that I need to do this:
$$\log_3((2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})+2^{64})=?$$
But afterwards, I can't find an algebric manipulation that will lead me to the solution, I took the exercise out of one of the entry tests of TAU.
Since \begin{align*} &\phantom{==}(3-2)(3+2)(3^2+ 2^2)(3^4+2^4)\cdots(3^{32}+2^{32})\\ &= (3^2-2^2)(3^2 + 2^2)(3^4+2^4)\cdots(3^{32}+2^{32})\\ &= (3^4-2^4)(3^4+2^4)(3^8+2^8)\cdots(3^{32}+2^{32})\\ &= \cdots\\ &= 3^{64}-2^{64}, \end{align*} the result is $\log_3(3^{64}) = 64$. The trick is $(a-b)(a+b)= a^2 - b^2$.