Question

Помогите пж с третьим заданием

Answer 1

$1) \ \text{log}_{0,9}(x-5) \geqslant \text{log}_{0,9}11$

$0,9 < 1 \Rightarrow y = \text{log}_{0,9} \downarrow \\\left \{ {\bigg{x-5>0 \ \ } \atop \bigg{x-5 \leqslant 11}} \right. \ \ \ \ \ \left \{ {\bigg{x>5 \ \ } \atop \bigg{x \leqslant 16}} \right.$

$\text{OTBET:} \ x \in (5; \ 16]$

$2) \ \log_{x}(x^{2} + 3x - 8) > 1\\\log_{x}(x^{2} + 3x - 8) > \log_{x}x$

$D(y): x^{2} + 3x - 8 > 0; x = -1,5 + \sqrt{41}; x \in (-1,5 + \sqrt{41}; + \infty)$

$\text{a)} \ x > 1 \Rightarrow y = \log_{x} \uparrow\\\begin{equation*} \begin{cases} x > 1, \\ x^{2} + 3x - 8 > x. \end{cases}\end{equation*}$

$\begin{equation*} \begin{cases}x > 1, \\ x_{1} = -4; \ x_{2} = 2. \end{cases}\end{equation*}$

$x \in (2; + \infty)$

При $x \in (-1,5 + \sqrt{41}; + \infty)$, то $x \in (2; + \infty)$

$\text{b)} \ x < 1 \Rightarrow y = \log_{x} \downarrow \\\begin{equation*} \begin{cases}x > 1, \\ x^{2} + 3x - 8 < x. \end{cases}\end{equation*}$

$\begin{equation*} \begin{cases}0<x<1, \\ x_{1} = -4; \ x_{2} = 2. \end{cases}\end{equation*}$

$x \in (0; 1)$

При $x \in (-1,5 + \sqrt{41}; + \infty)$, то $x \notin (0; 1)$

$\text{OTBET:} \ x \in (2; + \infty)$