Question

РЕШИТЕ ЛОГАРИФМИЧЕСКОЕ УРАВНЕНИЕ С ОДЗ​

Answer 1

$OD3: \left \{ {{\left \{ {{x-5>0} \atop {x+7>0}} \right. } \atop {x-5\neq 8}} \right. \left \{ {{x>5} \atop {x\neq 13}} \right.\Rightarrow x\in(5;8)\cup(8; +\infty)$$\lg\sqrt{x+7}-\lg2=\lg(x-5)-\lg8\\\\\lg \sqrt{x+7}-\lg(x-5)=\lg2-\lg8\\\\\lg(\frac{\sqrt{x+7} }{x-5} )=\lg\frac{2}{8}\\\\\frac{\sqrt{x+7}}{x-5}=\frac{2}{8};\\\\2(x-5)=8\sqrt{x+7};\\\\ 4(x-5)^2=64(x+7);\\\\4(x^2-10x+25)=64x+448\\\\4x^2-40x-64x+100-448=0;\\4x^2-104x-348=0|:4\\\\x^2-26x-87=0\Rightarrow x_1=-3\notin OD3, x_2=29.$

ОТВЕТ: 29.