Question

помогите решит неравенства


1 ($(frac{1}{32} )^2^x geq sqrt{0.125}$
2 log$^ frac{1}{3} (5x-1) geq 2$

Answer 1

1) (1/32)^(2x) >= sqrt(0.125)
0.5^(5*2x)>=0.5^(3/2)
10x <= 3/2
x<=3/20
x<=0.15
2) log(1/3, 5x-1) >=2
ОДЗ: 5x-1>0, x>0.2
log(1/3, 5x-1) >= log(1/3, 1/9)
5x-1 <= 1/9
x<=2/9
С учетом ОДЗ, 0.2 < x <= 2/9