Question

If 2 square x=3 square y=6 square-z, then show that 1/x+1/y+1/z=0

Answer 1

LET 2^x=3^y=(1/6)^z=k

Therefore,

K^(1/x)=2

K^(1/y)=3

K^(1/z)=(1/6)

Multiply these 3 equations to get:

K^[(1/x)+(1/y)+(1/z)]=1

Hence we can conclude that either k=1 ( which is totally false) or power is equal to “0”.

The correct answer is : [(1/x)+(1/y)+(1/z)]=0