Question

Помогите пожалуйста!!!
Решите неравенство:
log1/4(4x-3) ⩾-1

Answer 1

ОДЗ :

4x - 3 > 0

4x > 3

x > 0,75

$log_{\frac{1}{4}}(4x-3)\geq-1\\\\0<\frac{1}{4}<1 \Rightarrow 4x-3\leq (\frac{1}{4})^{-1}\\\\4x-3\leq 4\\\\4x\leq7\\\\x\leq1,75\\\\Otvet:\boxed{(0,75;1,75]}$

Answer 2

$log_{\frac{1}{4}}(4x-3)\geq -1\; \; ,\; \; \; ODZ:\; 4x-3>0\; ,\; \; x>\frac{3}{4}\; ,\; \; x>0,75\\\\a=\frac{1}{4}<1\; \; \Rightarrow \; \; \; 4x-3\leq (\frac{1}{4})^{-1}\\\\4x-3\leq 4\\\\4x\leq 7\\\\x\leq \frac{7}{4}\\\\x\leq 1,75\\\\Otvet:\; \; x\in (\; 0,75\, ;\, 1,75\; ]\; .$