Question

Помогите решить
(Log4((2^x)-1))/x-1<=1

Answer 1

$\dfrac{\log_4(2^x-1)}{x-1}\leq 1$

Область допустимых значений

$\displaystyle \left \{ {{2^x-1>0} \atop {x-1\neq 0}} \right. ~~\Leftrightarrow~~ \left \{ {{2^x>2^0} \atop {x\neq 1}} \right. ~~\Leftrightarrow~~ \left \{ {{x>0} \atop {x\neq 1}} \right.$

ОДЗ : x ∈ (0; 1) ∪ (1; +∞)

$1)~x\in (0;1); ~~(x-1)<0~~ \Rightarrow\\\\\dfrac{\log_4(2^x-1)}{x-1}\leq 1~~\Big|\cdot (x-1)<0\\\\\log_4(2^x-1)\geq x-1~~\Leftrightarrow~~\log_4(2^x-1)\geq \log_44^{x-1}\\\\2^x-1\geq 4^{x-1}~~|\cdot4~~~~\Leftrightarrow~~4\cdot 2^x-4\geq 4^x\\\\2^{2x}-4\cdot 2^x+4\leq 0\\\\(2^x-2)^2\leq 0;~~~2^x=2;~~~x=1;$

Не подходит  по ОДЗ

$2)~x\in (1;+\infty ); ~~(x-1)>0~~ \Rightarrow\\\\\dfrac{\log_4(2^x-1)}{x-1}\leq 1~~\Big|\cdot (x-1)>0\\\\\log_4(2^x-1)\leq x-1~~\Leftrightarrow~~\log_4(2^x-1)\leq \log_44^{x-1}\\\\2^x-1\leq 4^{x-1}~~|\cdot4~~~~\Leftrightarrow~~4\cdot 2^x-4\leq 4^x\\\\2^{2x}-4\cdot 2^x+4\geq 0\\\\(2^x-2)^2\geq 0;$

Квадрат выражения всегда неотрицательный.

Ответ: х ∈ (1; +∞)