Question

Solve the following equation:
$\log \left(64\sqrt[24]{242^{x^2-40x}}\right)=0$

Answer 1

$log\left(64\sqrt[24]{242^{x^2-40x}}\right)=0\Rightarrow 64.\sqrt[24]{242^{x^{2-40x}}}=1\Rightarrow (2^{x^2-40x}{)}^{\frac{1}{24}}=\frac{1}{64}\Rightarrow 2^{x^2-40x}={(2}^{-6}{)}^{24}\Rightarrow x^2-40x=-144\Rightarrow x^2-40x+144=0\Rightarrow (x-36\left)\right(x-4)=0\Rightarrow x=36,4.$