Question

log2(x+2)+log2(x-3)=log2(3+1)

Answer 1

ОДЗ :

$\left \{ {{x+2>0} \atop {x-3>0}} \right.\\\\\left \{ {{x>-2} \atop {x>3}} \right. \Rightarrow x\in(3;+\infty)$

$log_{2}(x+2)+log_{2}(x-3)=log_{2}(3+1)\\\\log_{2}(x+2)(x-3)=log_{2}4\\\\(x+2)(x-3)=4\\\\x^{2}-3x+2x-6-4=0\\\\x^{2}-x-10=0\\\\D=(-1)^{2}-4*(-10)=1+40=41\\\\x_{1}=\frac{1+\sqrt{41}}{2}\\\\x_{2}=\frac{1-\sqrt{41}}{2}-neyd\\\\Otvet:\frac{1+\sqrt{41}}{2}$