1.

8. If 2x:-3y6", show that-+-+--0

Answer»

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

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