Question

2 log x = log (2x- 3) + log (x + 2)

Answer 1

$2\log{x}=\log(2x-3)+\log(x+2)\qquad\qquad\qquad\begin{matrix}\text{Condi\c{c}\~ao de exist\^encia dos logaritmos:}\\\\\begin{matrix}2x-3>0\\x>\dfrac{3}{2}\end{matrix}\qquad\qquad\begin{matrix}x+2>0\\x>-2\end{matrix}\\\\\therefore\text{C.E.:}{x}>\dfrac{3}{2}\end{matrix}\\\\\log(x^2)=\log[(2x-3)(x+2)]\\\\\therefore{x}^2=(2x-3)(x+2)\\\\x^2=2x^2+x-6\\\\x^2+x-6=0\\\\(x+3)(x-2)=0\left\langle\begin{array}{lcl}x+3=0\longrightarrow{x}=-3\not\in\text{C.E.}\\\\x-2=0\longrightarrow{x}=2\in\text{C.E.}\end{array}\right\\\\\\S=\{2\}$