Пжл СрОчНО ЛОГАРИФМИЧЕСКИЕ УРАВНЕНИЯ!!!!!! Даю 50 баллов

Question

Дан 1 ответ

_zn · Answer 1 · 2020-05-26T03:50:52+0000

1. А) $\lg (x^{2} - 2x) - \lg (2x + 12) = 0$

ОДЗ: 0} \atop \bigg{2x + 12 > 0}} \right. \ \ \ \ \ \ \ \ \ \ \ \left \{ {\bigg{x \in (-\infty; 0)\cup(2; +\infty)} \atop \bigg{x \in (-6; +\infty) \ \ \ \ \ \ \ \ \ \ \ }} \right. \ \ \ \ \ \ \ \ x \in (-6; 0) \cup (2; +\infty)" alt="\left \{ {\bigg{x^{2} - 2x > 0} \atop \bigg{2x + 12 > 0}} \right. \ \ \ \ \ \ \ \ \ \ \ \left \{ {\bigg{x \in (-\infty; 0)\cup(2; +\infty)} \atop \bigg{x \in (-6; +\infty) \ \ \ \ \ \ \ \ \ \ \ }} \right. \ \ \ \ \ \ \ \ x \in (-6; 0) \cup (2; +\infty)" align="absmiddle" class="latex-formula">

$\lg (x^{2} - 2x) = \lg (2x + 12) \\x^{2} - 2x = 2x + 12\\x^{2} - 4x - 12 = 0\\x_{1} = -2; \ x_{2} = 6$

Ответ: $x_{1} = -2; \ x_{2} = 6$

Б) $\log_{2}(x+1) - \log_{2}(x-1) = 1$

ОДЗ: 0} \atop \bigg{x-1 > 0}} \right. \ \ \ \ \ \ \ \ \ \ \ \left \{ {\bigg{x \in (-1; +\infty)} \atop \bigg{x \in (1; +\infty) \ \ }} \right. \ \ \ \ \ \ \ \ x \in (1; +\infty)" alt="\left \{ {\bigg{x+1 > 0} \atop \bigg{x-1 > 0}} \right. \ \ \ \ \ \ \ \ \ \ \ \left \{ {\bigg{x \in (-1; +\infty)} \atop \bigg{x \in (1; +\infty) \ \ }} \right. \ \ \ \ \ \ \ \ x \in (1; +\infty)" align="absmiddle" class="latex-formula">

$\log_{2}(x+1) = \log_{2}2 + \log_{2}(x-1)\\\log_{2}(x+1) = \log_{2}2(x-1)\\x + 1 = 2(x-1)\\x + 1 = 2x - 2\\x = 3$

Ответ: $x = 3$

В) $2\log_{4}^{2}x + 5\log_{4}x - 3 = 0$

ОДЗ: 0; \ x \in (0; + \infty)" alt="x > 0; \ x \in (0; + \infty)" align="absmiddle" class="latex-formula">

Замена: $\log_{4}x = t$

$2t^{2} + 5t - 3 = 0\\D = 25 + 24 = 49\\\\t_{1,2} = \dfrac{-5 \pm 7}{4} = \left[\begin{array}{ccc}t_{1} = -3 \\t_{2} = \dfrac{1}{2} \end{array}\right$

$1) \ \log_{4}x = -3; \ x = 4^{-3} = \dfrac{1}{64} \\2) \ \log_{4}x = \dfrac{1}{2}; \ x = \sqrt{4} = 2$

Ответ: $x_{1} = \dfrac{1}{64}; \ x_{2} = 2$

Г) $x^{\log _{0,5}(x)-1} = \dfrac{1}{64}$

ОДЗ: 0; \ x \in (0; + \infty)" alt="x > 0; \ x \in (0; + \infty)" align="absmiddle" class="latex-formula">

$\log _{0,5}x^{\log _{0,5}(x)-1} = \log _{0,5}\bigg(\dfrac{1}{64}\bigg)\\(\log _{0,5}(x)-1) \ \cdotp \log _{0,5}x = 6\\\log_{0,5}^{2}x - \log _{0,5}x - 6 = 0$

Замена: $\log _{0,5}x = t$

$t^{2} - t - 6 = 0; \\t_{1} = -2; \ t_{2} = 3$

$1) \ \log _{0,5}x = -2; \ x = 0,5^{-2} = 4\\2) \ \log _{0,5}x =3; \ x = 0,5^{3} = 0,125$

Ответ: $x = 0,125; \ x = 4$

2. $\left \{ {\bigg{\log_{2}(x^{2} + 4xy + 4y^{2}) = 4} \atop \bigg{\log_{2}x=\log_{2}(2y - 4) \ \ \ \ \ }} \right. \ \ \ \ \ \ \ \ \left \{ {\bigg{\log_{2}(x+2y)^{2} = 4} \atop \bigg{x=2y - 4 \ \ \ \ \ \ \ \ \ }} \right.$

$\left \{ {\bigg{\log_{2}(x+2y) = 2} \atop \bigg{x=2y - 4 \ \ \ \ \ \ \ }} \right.\\\log_{2}(2y - 4+2y) = 2\\\log_{2}(4y - 4) = \log_{2}4\\4y - 4 = 4\\4y = 8\\y = 2\\x=2 \ \cdotp 2 - 4 = 0$

Ответ: $(0; 2)$