Question

Solve the following equation:
$\frac{{\log }_2(9-2^x)}{3-x}=1$

Answer 1

${\log }_2(9-2^x)=3-x9-2^x=2^{3-x}9=2^x+2^{3-x}9=2^x+\frac{8}{2^x}\Rightarrow {\left(2^x\right)}^2-9\cdot 2^x+8=02^x=\frac{9\pm \sqrt{81-4\times 8}}{2}=\frac{9\pm 7}{2}=1,8$

$2^x=1x=0$ and $2^x=8x=3$