Question

Solve the following equation:
$\log (3^x-2^{4-x})=2+\frac{1}{4}\log 16-\frac{x\log 4}{2}$

Answer 1

$\log (3^x-2^{4-x})=2+\frac{1}{4}\log 16-\frac{x\log 4}{2}\Rightarrow \log (3^x-2^{4-x})=2+\log 2-\log 4^{\left(\frac{x}{2}\right)}\Rightarrow \log (3^x-2^{4-x})=\log \left(\frac{200}{2x}\right)\Rightarrow 3^x-2^{4-x}=\frac{200}{2^x}\Rightarrow 6^x-2^4=200\Rightarrow 6^x=216$

Therefore, $x=3.$