Solve the inequality $\log_3(2^{x+1}-8)\le x$ Find x
$$\log_3(2^{x+1}-8)≤x$$
I found that $x>2$
$$\log_3(2^{x+1}-8)\le x\log_33$$
$$(2^{x+1}-8)\le 3^x$$
$$(2^{x+1}-8)> 0$$
$$(2^{x+1})>8$$
$$2^{x+1}>2^3$$
$${x+1}>3$$
$${x}>2$$

We have
$$\log_3(2^{x+1}-8)≤x$$
$$\log_3(2^{x+1}-8)≤\log_33^x$$
$$\begin{align}&\begin{cases} 2^{x+1}-3^x-8≤0 \\x+1>3 \end{cases} \\\\ \iff &\begin{cases}3^x-2^{x+1}≥-8 \\ x>2\end{cases}\end{align}$$
Then, note that  $f(x)$ is a strictly increasing function, $\forall x>2.$
This means,
$$f(x)=3^x-2^{x+1}>1≥-8,  \forall x>2$$
Finally, you get the right answer
$$x\in(2, +\infty)$$

