Алгебра: Решите логарифмическое неравенство

Объяснение:ОДЗ: -∞__+__3__-__7__+__+∞x∈(-∞;3)U(7;+∞).1.1.2.ответ: x∈(-∞;3)U[7¹/₂₇;+∞)....

Решите логарифмическое неравенство

Ответ:

nshexovceva18

25.09.2021 23:20

Объяснение:

$log_9(x-7)^2*log_{81}(x-3)^4+log_3\frac{(x-3)^3}{x-7}\geq 3.$

ОДЗ:

$\frac{(x-3)^3}{x-7}0\ \ \ \ \frac{x-3}{x-7}0.$

-∞__+__3__-__7__+__+∞

x∈(-∞;3)U(7;+∞).

$2*log_{3^2}(x-7)*4*log_{3^4}(x-3)+log_3(x-3)^3-log_3(x-7)-3\geq 0\\\frac{2}{2}*log_3(x-7)*\frac{4}{4} *log_3(x-3)+3*log_3(x-3)-log_3(x-7)-3\geq 0\\log_3(x-7)*log_3(x-3)+3*log_3(x-3)-log_3(x-7)-3\geq 0\\ log_3(x-7)*log_3(x-3)-log_3(x-7)+3*log_3(x-3)-3\geq 0\\log_3(x-7)*(log_3(x-3)-1)+3*(log_3(x-3)-1)\geq 0\\(log_3(x-3) -1)*(log_3(x-7)+3)\geq 0.$

$\left \{ {{log_3(x-3)-1\geq 0} \atop {log_3(x-7)+3\geq 0}} \right. \ \ \ \ \left \{ {log_3(x-3)\geq 1} \atop {log_3(x-7)\geq -3}} \right. \ \ \ \ \left \{ {{x-3\geq 3^1} \atop {x-7\geq 3^{-3}}} \right. \ \ \ \ \left \{ {{x\geq 6} \atop {x-7\geq \frac{1}{27} }} \right. \ \ \ \ \left \{ {{x\geq 6} \atop {x\geq 7\frac{1}{27} }} \right. \ \ \ \ \Rightarrow\\x\in[7\frac{1}{27} +\infty).$

$\left \{ {{log_3(x-3)-1\leq 0} \atop {log_3(x-7)+3\leq 0}} \right. \ \ \ \ \left \{ {log_3(x-3)\leq 1} \atop {log_3(x-7)\leq -3}} \right. \ \ \ \ \left \{ {{x-3\leq 3^1} \atop {x-7\leq 3^{-3}}} \right. \ \ \ \ \left \{ {{x\leq 6} \atop {x-7\leq \frac{1}{27} }} \right. \ \ \ \ \left \{ {{x\leq 6} \atop {x\leq 7\frac{1}{27} }} \right. \ \ \ \ \Rightarrow\\x\in(-\infty;6].\ \ \ \ \Rightarrow$