Solve for: $2\log_3\left(x^2-4\right)+3\sqrt{\log_3\left(x+2\right)^2}-\log_3\left(x-2\right)^2\leq4$

Question

Solve for: $$2\log_3\left(x^2-4\right)+3\sqrt{\log_3\left(x+2\right)^2}-\log_3\left(x-2\right)^2\leq4$$

My try:

$2\log_3\left(x^2-4\right)+3\sqrt{\log_3\left(x+2\right)^2}-\log_3\left(x-2\right)^2\leq4\\\Leftrightarrow \log_3\left(x^2-4\right)^2+3\sqrt{\log_3\left(x+2\right)^2}-\log_3\left(x-2\right)^2\leq4\\\Leftrightarrow \log_3\left[\left(x-2\right)^2\times\left(x+2\right)^2\right]+3\sqrt{\log_3\left(x+2\right)^2}-\log_3\left(x-2\right)^2\leq4\\\Leftrightarrow \log_3\left(x-2\right)^2+\log_3\left(x+2\right)^2+3\sqrt{\log_3\left(x+2\right)^2}-\log_3\left(x-2\right)^2\leq4\\\Leftrightarrow \log_3\left(x+2\right)^2+3\sqrt{\log_3\left(x+2\right)^2}-4\leq0\,\,\,(*)$

Put: $t=\sqrt{\log_3\left(x+2\right)^2}\Rightarrow (*)\Leftrightarrow t^2+3t-4\leq0$

But I don't know Conditions defined for this math? Could help me?

Answer 1

The domain of the first inequation is $x>3$ or $x\leq 3$, so the range of your $t$ is $t\geq 0$. Then you can use $t\geq 0$ as the domain to solve $t^2+3t−4\leq0$, which gets $0\leq t \leq 1$. At last, you just go back to solve $0 \leq t \leq 1$.