${\log }_2\left(5\cdot 2^x+1\right), \frac{1}{2}{\log }_2\left(2^{1-x}+1\right), {\log }_{2}2$ are in A.P.
$\Rightarrow 2\times \frac{1}{2}{\log }_2\left(2^{1-x}+1\right)={\log }_2\left(5\cdot 2^x+1\right)+\mathrm{l}\mathrm{o}{\mathrm{g}}_{2}2$
$\Rightarrow 2^{1-x}+1=\left(5\cdot 2^x+1\right)2$
Let, $2^x=y$

$\Rightarrow \frac{2}{y}+1=\left(5y+1\right)2$
$\Rightarrow 2+y=10y^2+2y$
$\Rightarrow 10y^2+y-2=0\Rightarrow y=\frac{2}{5},\frac{-1}{2}$
$\Rightarrow 2^x=\frac{2}{5}$ or $\frac{-1}{2}$(not possible)

$\Rightarrow 2^x=\frac{2}{5}$
$\Rightarrow x={\mathrm{l}\mathrm{o}\mathrm{g}}_2\left(\frac{2}{5}\right)=1-{\log }_{2}5=1-\frac{\log 5}{\log 2}$