Question

64^x=48^y=36^z,prove,1/x+1/z=2/y

Answer 1

Let (64)^x =(48)^y =(36)^z =K

So 64= K^1/x

  48 =K^1/y

  36 = K^1/z

So now K^1/y / K^1/x = 48/64

Or K^(1/y-1/x) =48/64 =3/4

AGAIN SIMILARLY

K^(1/z-1/y) =36/48 =3/4

SO K^(1/y -1/x) = K^(1/z -1/y) =3/4

Or (1/y -1/x)= (1/z -1/y) as base "K" is samefor bothe index.

Or1/y + 1/y = 1/z + 1/x

Or 2/y = 1/z + 1/x

So 1/x + 1/z =2/y..PROVED.

 

 

OrK^{1/2(1/z-1/x)} =3/4

Answer 2

Step-by-step explanation:

We know that

64×36=(48)^2

Let

(64)^x=(48)^y=(36)^z=k

or; 64=k^1/x

48=k^1/y

36=k^1/z

64×36=(48)^2

k^1/x × k^1/y = (k^1/z)^2

or, 1/x + 1/y = 2/z [Proved ]