<s>[INST] 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.
 [/INST] We have
$$3-2\cdot 3^{x+1} = 3^{2+2x}$$
Setting $3^{x+1}=t$, we obtain
$$3-2t = t^2 \implies t^2+2t-3 =0 \implies (t+3)(t-1) = 0 \implies t \in \{-3,1\}$$
Hence, if we want $x \in \mathbb{R}$, we have
$$3^{x+1} = 1 \implies x+1 =0 \implies x=-1$$
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