Log3(8x-3)≥0 F(x)=1/3x2-1/2X2+7

Log₃(8x-3)≥0, 0=log₃3⁰=l0g₃1

log₃(8x-3)≥log₃1
основание логарифма а=3, 3>1. знак неравенства не меняем:
$\left \{ {{8x-3\ \textgreater \ 0} \atop {8x-3} \geq 1 \right. , \left \{ {{8x\ \textgreater \ 3} \atop {8x \geq 4}} \right. , \left \{ {{x\ \textgreater \ \frac{3}{8} } \atop {x \geq \frac{4}{8} }} \right. =\ \textgreater \ x \geq \frac{4}{8}$
x≥0,5
ответ: x∈[0,5;∞)