решить неравенство log(2-0.1x) по основанию 9/16 < 1/2

$log_{\frac{9}{16}} (2-0.1x)<\frac{1}{2}$

0" alt="2-0.1x>0" align="absmiddle" class="latex-formula">
$0<\frac{9}{16}<1$ ;

(\frac{9}{16})^{\frac{1}{2}}" alt="2-0.1x>(\frac{9}{16})^{\frac{1}{2}}" align="absmiddle" class="latex-formula">

\sqrt{\frac{9}{16}}=\sqrt{(\frac{3}{4})^2}=\frac{3}{4}=0.75" alt="2-0.1x>\sqrt{\frac{9}{16}}=\sqrt{(\frac{3}{4})^2}=\frac{3}{4}=0.75" align="absmiddle" class="latex-formula">

-2+0.75" alt="-0.1x>-2+0.75" align="absmiddle" class="latex-formula">

-1.25" alt="-0.1x>-1.25" align="absmiddle" class="latex-formula"> |:-0.1<0<br> $x<12.5$
x є $(-\infty;12.5)$