Q) If 2^x = 3^y = 6^z , then Prove that 1/z = 1/x + 1/y .

A) 2^x=3^y=6^z
3logy = zlog6
y = zlog6/log3

xlog2 = ylog3
xlog2 = zlog6log3/log3
x = zlog6/log2

1/x + 1/y = log3/zlog6 + log2/zlog6
= log(3*2)/zlog6
= log6/zlog6
=1/z

so 1/x + 1/y = 1/z

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