Question

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

Answer 1

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

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

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

$\Rightarrow 4^x+4=2\cdot 4^x-3\cdot 2^x$

$\Rightarrow 4^x-3\times 2^x-4=0$

$\Rightarrow 2^x=\frac{+3\pm \sqrt{9+16}}{2}=\frac{3\pm 5}{2}=4,-1$

$\Rightarrow 2^x=4$

$\Rightarrow x=2$

$\Rightarrow 2^x=-1$

$\Rightarrow x=imaginary$