Решите показательное неравенство [latex]3 ^{2x} -3 ^{x} \leq 72[/latex]

[latex]3^{2x} - 3^x - 72 \leq 0\\ 3^x = t, \ \ t \ \textgreater \ 0 \\ t^2 - t - 72 \leq 0 \\ t_{12} = \frac{1б\sqrt{1+288}}{2} = \frac{1б17}{2} \\ -8 \leq t \leq 9 \\ t \ \textgreater \ 0 =\ \textgreater \ \ \ \ t \leq 9 \\ 3^x \leq 3^2 \\ 3 \ \textgreater \ 1 =\ \textgreater \ \ \ \ \ x \leq 2[/latex]