Solve this equation: $\log_3(3-2\cdot3^{x+1})=2+2x$

Question

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.

Answer 1

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 $$3^{x+1} = 1 \implies x+1 =0 \implies x=-1$$

Answer 2

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 $x=-1$.

Answer 3

$$3-2\cdot 3^{x+1} = (3^{x+1})^2$$

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 $$x = \log_3(1) - 1 = 0 - 1 = -1.$$