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.

$$3-2\cdot 3^{x+1} = (3^{x+1})^2$$
Set $z = 3^{x+1}$:
$$z^2 + 2\cdot z-3 = 0$$
Solve quadratic equation:
$$z = -1 \pm \sqrt{1 + 3} = -1 \pm 2$$
so $z \in \{-3,1\}$.
Compute $x$ from
$$x = \log_3(z) - 1.$$
For a real solution, you have to pick $z=1$, so
$$x = \log_3(1) - 1 = 0 - 1 = -1.$$
