<s>[INST] Evaluating $\log_3(1+2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1))$ in the most efficient way I have come across a tricky question while studying logarithms.
$$\log_3(1+2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1))$$
While plugging it into a calculator brings a seemingly simple answer, I cannot find a way to start this without a calculator.
 [/INST] Rewrite $\log_3(1+2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1))$ as $\log_3(1+(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1))$. You can use the Difference of Squares formula to find $\log_3(1+(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)(3^{32}+1))$. If you use the Difference of Squares formula 5 times, you get $\log_3(1+(3^{64}-1))$. Finally, by removing the parenthesis, you get $\log_3(3^{64}+1-1)$, which simplifies to $\log_3(3^{64})$. Now, with the logarithm power rule, this becomes $64\log_3(3)$, which is equal to $64$.
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