Question

2^4-5х>1/32
2^4-5х>2^-5
4-5х>-5
9>5х
Х<9:5
Х€(-бесконечность;9/5)
Подробно т.к напишите пжлст

Answer 1

$2^{4-5x}\ \textgreater \ \frac{1}{32}\\\\2^{4-5x}\ \textgreater \ 32^{-1}\\\\2^{4-5x}\ \textgreater \ 2^{-5}\\\\4-5x\ \textgreater \ -5\\-5x\ \textgreater \ -9\\5x\ \textless \ 9\\x\ \textless \ \frac{9}5\\x\in(-\infty;+\frac{9}5)$