Question

Log3(8x-3)≥0 F(x)=1/3x2-1/2X2+7

Answer 1

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;∞)