Question

Solve the following equation:
$\log 2+\log (4^{x-2}+9)=1+\log (2^{x-2}+1)$

Answer 1

$\log 2+\log (4^{x-2}+9)=1+\log (2^{x-2}+1)$

$\Rightarrow 2.(4^{x-2}+9)=10.(2^{x-2}+1)\Rightarrow 2.{{(2}^{x-2})}^2-10.{(2}^{x-2})+8=0\Rightarrow {\left(2^{x-2}\right)}^2-5(2^{x-2})+4=0\Rightarrow 2^{x-2}=1,4\Rightarrow x-2=0,2\Rightarrow x=2,4$