Question

Логарифмическое уравнение.

Answer 1

ОДЗ : x > 0 , x ≠ 1/3

$log_{3x} \frac{3}{x}+log_{3} ^{2}x=1\\\\\frac{log_{3} \frac{3}{x}}{log_{3}3x }+log_{3}^{2}x=1\\\\\frac{log_{3}3-log_{3}x }{log_{3}3+log_{3}x} +log_{3}^{2}x=1\\\\\frac{1-log_{3}x }{1+log_{3}x }+log_{3}^{2}x=1\\\\log_{3} x=m,m\neq-1$

$\frac{1-m}{1+m}+m^{2}-1=0\\\\\frac{1-m+m^{2} +m^{3}-1-m }{1+m}=0\\\\\left \{ {{m^{3}+m^{2} -2m=0 } \atop {1+m\neq0 }} \right.\\\\m^{3}+m^{2}-2m=0\\\\m(m^{2}+m-2)=0\\\\m(m+2)(m-1)=0\\\\m_{1}=0\\\\m_{2}=-2\\\\m_{3}=1\\\\log_{3}x=0\\\\x_{1}=1\\\\log_{3}x=-2\\\\x_{2} =\frac{1}{9} \\\\log_{3}x=1\\\\x_{3}=3\\\\Otvet:\boxed{1;\frac{1}{9};3}$

Answer 2

Ответ: во вложении Объяснение: