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Question Number 74589 by Aditya789 last updated on 27/Nov/19

Answered by MJS last updated on 27/Nov/19

a(b−c)x^2 +b(c−a)xy+c(a−b)y^2 =0 x^2 +((b(c−a)y)/(a(b−c)))x+((c(a−b)y^2 )/(a(b−c)))=0 x=t−((b(c−a)y)/(2a(b−c))) t^2 −((((ab−2ac+bc)y)/(2a(b−c))))^2 =0 to get a perfect square t=0 ⇒ y=0∨b=((2ac)/(a+c)) b=((2ac)/(a+c)) ⇒ ((log (a+c) +log (a−2b+c))/(log (a−c)))= =((log (a+c) +log (((a−c)^2 )/(a+c)))/(log (a−c)))= =((log (a+c) +2log (a−c) −log (a+c))/(log (a−c)))=2

a(b−c)x2+b(c−a)xy+c(a−b)y2=0x2+b(c−a)ya(b−c)x+c(a−b)y2a(b−c)=0x=t−b(c−a)y2a(b−c)t2−((ab−2ac+bc)y2a(b−c))2=0togetaperfectsquaret=0⇒y=0∨b=2aca+cb=2aca+c⇒log(a+c)+log(a−2b+c)log(a−c)==log(a+c)+log(a−c)2a+clog(a−c)==log(a+c)+2log(a−c)−log(a+c)log(a−c)=2

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