Question

Solve the following equation:
${\log }_2(4^x+1)=x+{\log }_2(2^{x+3}-6)$

Answer 1

${\log }_2(4^x+1)=x+{\log }_2(2^{x+3}-6)\Rightarrow {\log }_2(4^x+1)={\log }_2{2}^x+{\log }_2(2^{x+3}-6)\Rightarrow 4^x+1=2^x(2^{x+3}-6)\Rightarrow 4^x+1={8.4}^x-{6.2}^x\Rightarrow {7.4}^x-{6.2}^x-1=0\Rightarrow 2^x=1,-\frac{1}{7}$

But $2^x$ never be negative.

Therefore $2^x=1\Rightarrow x=0$