Răspuns
{4}
Explicație pas cu pas:
[tex]2log_{a}(x-2y)=log_{a}x+log_{a}y\\log_{a}(x-2y)^{2}=log_{a}(xy)\\x^{2}-4xy+4y^{2}=x*y\\x^{2}-5xy+4y^{2} =0\\\\\frac{x^{2}}{y^{2}} -5\frac{x}{y} +4=0\\\\t^{2} -5t+4=0\\(t-1)(t-4)=0\\t_{1}=1\\t_{2}=4\\[/tex]
Dar x>0,y>0
x-2y>0
x>2y
[tex]\frac{x}{y} \geq 2[/tex]
