Question

rezolvati in multimea numerelor reale ecuatia log _x din (2x+4)=log _x din (x^2+1)
Am cautat ore intregi o rezolvare si nu gasesc.Nu imi pot da seama cum se rezolva.

Answer 1

$log_{_x}(2x+4)=log_{_x}(x^2+1)\\\\\implies 2x+4=x^2+1\\x^2-2x-4+1=0\\x^2-2x-3=0\\\\{\Delta}=b^2-4ac=(-2)^2-4[1*(-3)]=4+12=16\\\\x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+\sqrt16}{2*1}=\frac{2+4}{2}=\frac{6}{2}=3\\\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-\sqrt16}{2*1}=\frac{2-4}{2}=\frac{-2}{2}=-1\\\\S:\ x\in(-1;3)$