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8. If 2x:-3y6", show that-+-+--0 |
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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 Like my answer if you find it useful! |
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