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