\frac{4}{3} " alt="log_{2}(3-2x)-log_{2^{-3}}(3-2x)> \frac{4}{3} " align="absmiddle" class="latex-formula">
\frac{4}{3} " alt="log_{2}(3-2x)+log_{2}(3-2x)^{ \frac{1}{3}} > \frac{4}{3} " align="absmiddle" class="latex-formula">
\frac{4}{3}" alt="log_{2}(3-2x)^{ \frac{4}{3}} > \frac{4}{3}" align="absmiddle" class="latex-formula">
\frac{4}{3}" alt="\frac{4}{3}*log_{2}(3-2x)>\frac{4}{3}" align="absmiddle" class="latex-formula">
1" alt="log_{2}(3-2x)>1" align="absmiddle" class="latex-formula">
2" alt="3-2x>2" align="absmiddle" class="latex-formula">
2-3" alt="-2x>2-3" align="absmiddle" class="latex-formula">
-1" alt="-2x>-1" align="absmiddle" class="latex-formula">
Ответ: x∈(-бесконечность; 0.5)