Question

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

Answer 1

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

ОДЗ: $left { {bigg{x^{2} - 2x > 0} atop bigg{2x + 12 > 0}} right. left { {bigg{x in (-infty; 0)cup(2; +infty)} atop bigg{x in (-6; +infty) }} right. x in (-6; 0) cup (2; +infty)$

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

Ответ: $x_{1} = -2; x_{2} = 6$

Б) $log_{2}(x+1) - log_{2}(x-1) = 1$

ОДЗ: $left { {bigg{x+1 > 0} atop bigg{x-1 > 0}} right. left { {bigg{x in (-1; +infty)} atop bigg{x in (1; +infty) }} right. x in (1; +infty)$

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

Ответ: $x = 3$

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

ОДЗ: $x > 0; x in (0; + infty)$

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: $x > 0; x in (0; + infty)$

$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: $(0; 2)$