Question

Solve the following equation.
${\log }_2(4^x+4)={\log }_2{2}^x+{\log }_2(2^{x+1}-3)$

Answer 1

${\log }_2(4^x+4)={\log }_2{2}^x+{\log }_2(2^{x+1}-3)\Rightarrow (4^x+4)=2^x(2^{x+1}-3)\Rightarrow 4^x-{3.2}^x-4=0$

But $2^x$ cannot be $<0.$

Therefore, $2^x=4\Rightarrow x=2.$