For A.P.: $2\log_{4}(2^{1-x}+1)=\log_{2}(5\cdot2^{x}+1)+1$. Since $\log_{4}y=\tfrac12\log_{2}y$, $\log_{2}(2^{1-x}+1)=\log_{2}\!\big(2(5\cdot2^{x}+1)\big)$. Thus $2^{1-x}+1=10\cdot2^{x}+2$. Let $t=2^{x}>0$. Then $\frac{2}{t}+1=10t+2 \Rightarrow 10t^{2}+t-2=0$. $t=\frac{-1+9}{20}=\frac{2}{5}$ (positive root). Hence $2^{x}=\frac{2}{5}$, so $x=\log_{2}\!\left(\frac{2}{5}\right)=1-\log_{2}5$.