Question Number 101225 by harckinwunmy last updated on 01/Jul/20

Commented byDwaipayan Shikari last updated on 01/Jul/20

x^a =y^b =z^c =k(k≠0)  x=k^(1/a)  y=k^(1/b)    z=k^(1/c)   y^2 =k^(2/b)   xz=k^((1/a)+(1/c))   (2/b)=(1/a)+(1/c)⇒((2a−b)/(ab))=(1/c)⇒c=((ab)/(2a−b))

Commented bysmridha last updated on 01/Jul/20

2.((log(x))/((y−z)))=((log(y))/((z−x)))=((log(z))/((x−y)))=u  now   x^x .y^y .z^z =e^(xlog(x)+ylog(y)+zlog(z))   putting the values  we get         =e^(ux(y−z)+uy(z−x)+uz(x−y))          =e^0 =1

Commented bybemath last updated on 01/Jul/20

(3a)(√(p+(√q))) = (√x) + (√y)  (√(((√x)+(√y))^2 )) = (√(x+y+2(√(xy))))  → { ((p=x+y)),((q=4xy)) :}

Commented byDwaipayan Shikari last updated on 01/Jul/20

p+(√q)=x+y+2(√(xy))  p=x+y  (√q)=2(√(xy))  (p^2 −q)^(1/2) =((x+y)^2 −4xy)^(1/2) =x−y