Finding value of $x$ in an A.P. If $1$ , $\log_{9}(3^{x+1} + 2)$, $\log_{3}(4⋅3^{x}-1)$ are in A.P. , then $x$ equals ?

$$2\log_{9}(3^{x+1} + 2)=1+\log_{3}(4⋅3^{x}-1)$$
We have
$$\log_{3}(3^{x+1} + 2)=\log_{3}(4⋅3^{x+1}-3)$$
therefore
$$3^{x+1} + 2=4⋅3^{x+1}-3\implies 3^{x+2}=5$$
$$x=-2+\log_{3}(5)$$
