If $1,\ \log_{9}!\left(3^{,1-x}+2\right)$ and $\log_{3}!\left(4\cdot 3^{x}-1\right)$ are in A.P., then $x$ equals

A.P. $\Rightarrow 2\cdot\log_{9}(3^{1-x}+2)=1+\log_{3}(4\cdot 3^x-1)$. Since $\log_{9}y=\dfrac12\log_{3}y$, we get $\log_{3}(3^{1-x}+2)=1+\log_{3}(4\cdot 3^x-1)=\log_{3}!\big(3(4\cdot 3^x-1)\big)$. Hence $3^{1-x}+2=12\cdot 3^{x}-3$. Put $t=3^x$: $\dfrac{3}{t}+2=12t-3$. $\Rightarrow 12t^2-5t-3=0 \Rightarrow t=\dfrac{3}{4}$ (positive root). So $3^x=\dfrac{3}{4}\Rightarrow x=\log_{3}!\left(\dfrac{3}{4}\right)=1-\log_{3}4$.