2^x-1≥8 log0.2(2x+3)≥-3 решите неравенства

[latex]2^x - 1 \geq 8 \\ 2^x \geq 9 \\ 2^x \geq 2^{log_29} \\ x \geq log_29[/latex] [latex]2x + 3 \ \textgreater \ 0 \\ x\ \textgreater \ - \frac{3}{2} \\ log_{ \frac{1}{5} }(2x -3) \geq -3 \\ log_5(2x-3) \leq 3 \\ log_5(2x-3) \leq log_5125 \\ 2x + 3 \leq 125 \\ 2x \leq 122 \\ x \leq 61[/latex] x ∈ [latex](- \frac{3}{2} ;61][/latex]