Question

log0,3(x+1)/log0,3⁡100- log0,3⁡9 <1

Answer 1

$\frac{\log_{0.3}(x+1)}{\log_{0.3}100-\log_{0.3}9} \ \textless \ 1\\ \\ \log_{0.3}(x+1)\ \textless \ \log_{0.3}100-\log_{0.3}9\\ \\ \log_{0.3}(x+1)\ \textless \ \log_{0.3} \frac{100}{9} \\ \\x+1\ \textgreater \ \frac{100}{9} \\ \\ x\ \textgreater \ \frac{91}{9}$
ОДЗ: $x> -1$

$-1\ \textless \ x\ \textless \ \frac{91}{9}$ - ответ