Question

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

Answer 1

$log_{\frac{9}{16}} (2-0.1x)<\frac{1}{2}$
$2-0.1x0$
$0<\frac{9}{16}<1$; $2-0.1x(\frac{9}{16})^{\frac{1}{2}}$
$2-0.1x\sqrt{\frac{9}{16}}=\sqrt{(\frac{3}{4})^2}=\frac{3}{4}=0.75$
$-0.1x-2+0.75$
$-0.1x-1.25$  |:-0.1<0
$x<12.5$
x є $(-\infty;12.5)$