Question

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

Answer 1

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