solve $x-\log_3 (9-3^x) \geq -2 $

Question

$x-\log_3 (9-3^x) \geq -2 $

original problem $ \frac{x-1}{\log_3 (9-3^x) -3} \leq 1$

so I solved this original problem till I got stuck there. In my book, I also found a way to solve this by converting 3 into $\log_3 27 $ but I found it non-intuitive. How am I supposed to know that?!! Is there a better way?

Answer 1

Hint:

$x+2\ge \log_3(9-3^x)$

$3^{x+2}\ge 9-3^x$

$9\cdot 3^x+3^x\ge 9$

Answer 2

Hint: Consider $$3^{x-log_3(9-3^x)} \geq 3^{-2},$$ and set $3^x=A.$