Question

решить уравнение
log2(x^2-3x + 2)=log2(2x-4)​

Answer 1

$log_2(x^2-3x+2)=log_2(2x-4)\\\\ODZ:\ \left\{\begin{array}{l}x^2-3x+2>0\\2x-4>0\end{array}\right\ \ \left\{\begin{array}{l}(x-1)(x-2)>0\\2x>4\end{array}\right\ \ \left\{\begin{array}{l}x\in (-\infty ,1)\cup (2,+\infty )\\x>2\end{array}\right\\\\\\x\in (2,+\infty )\\\\x^2-3x+2=2x-4\ \ ,\\\\x^2-5x+6=0\ \ \to \ \ x_1=2\notin ODZ\ \ ,\ \ x_2=3\ \ (teorema\ Vieta)\\\\Otvet:\ \ x=3\ .$