Solve this equation: $\log_3(3-2\cdot3^{x+1})=2+2x$ Solve this equation: $\log_3(3-2\cdot3^{x+1})=2+2x$. I put $(2+2x)^3=3-2\cdot3^{x+1}$. But I don't know how to go on.

Hint: $3-2\cdot 3^{x+1}>0$ 
So we'll have: $\:3^{2+2x}+2\cdot 3^{x+1}=3\:\:\rightarrow 9\cdot 3^{2x}+6\cdot 3^x=3\:\:\rightarrow \:3\cdot 3^{2x}+2\cdot 3^x-1=0\:$
let $3^x=a\:\:\rightarrow \:3a^2+2a-1=0$ with $a_1=-1$ and $a_2=\frac{1}{3}$, and only $a_2=\frac{1}{3}$ is good. Therefore, comeback at substitution we'll obtain: $3^x=\frac{1}{3}$ with only solution $x=-1$.
