If $1,{\log }_9\left(3^{1-x}+2\right),{\log }_3\left(4\cdot 3^x-1\right)$ are in AP, then xis equal to

Given,  $1,{\log }_9\left(3^{1-x}+2\right),{\log }_3\left(4\cdot 3^x-1\right)$ are in AP

$\begin{array}{l}\Rightarrow {\log }_{3}3,{\log }_{3^2}\left(3^{1-x}+2\right),{\log }_3\left(4\cdot 3^x-1\right)\text{ are in AP. }\\ \Rightarrow {\log }_{3}3\cdot \frac{1}{2}{\log }_3\left(3^{1-x}+2\right),{\log }_3\left(4\cdot 3^x-1\right)\text{ are in AP. }\\ \Rightarrow {\log }_3\left(3^{1-x}+2\right)={\log }_{3}3+{\log }_3\left(4\cdot 3^x-1\right)\\ \Rightarrow {\log }_3\left(3^{1-x}+2\right)={\log }_{3}3\cdot \left(4\cdot 3^x-1\right)\\ \Rightarrow 3^{1-x}+2=3\cdot \left(4\cdot 3^x-1\right)\\ \Rightarrow 3^{1-x}+2=12\cdot 3^x-3\\ \Rightarrow 3+2\cdot 3^x=12{\left(3^x\right)}^2-3\cdot 3^x\\ \Rightarrow 12{\left(3^x\right)}^2-5\left(3^x\right)-3=0\end{array}$

Put $3^x=t$, then

$12t^2-5t-3=0$

$\therefore$  $t=\frac{5\pm \sqrt{25+144}}{24}$

$=\frac{5\pm \sqrt{169}}{24}=\frac{5\pm 13}{24}=\frac{3}{4},-\frac{1}{3}=$

$\left[\because t=3^x=-\frac{1}{3}\right.$ s not possible, because $\left.3^x>0,\mathrm{\forall }x\right]$

When $t=\frac{3}{4}$,. then

$\begin{array}{l}{3}^x=\frac{3}{4}\\ \Rightarrow x={\log }_3(3/4)=1-{\log }_{3}4\end{array}$