Question

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

Answer 1

$2^x - 1 geq 8 \ 2^x geq 9 \ 2^x geq 2^{log_29} \ x geq log_29$

$2x + 3 textgreater 0 \ x textgreater - frac{3}{2} \ log_{ frac{1}{5} }(2x -3) geq -3 \ log_5(2x-3) leq 3 \ log_5(2x-3) leq log_5125 \ 2x + 3 leq 125 \ 2x leq 122 \ x leq 61$

x ∈ $(- frac{3}{2} ;61]$