If 2^x = 3^y = 6^z , then (1/x) + (1/y) + (1/z) = ___


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Question Number 62591 by hovea cw last updated on 23/Jun/19

$If 2^{x} = 3^{y} = 6^{z}, then \frac{1}{x} + \frac{1}{y} + \frac{1}{z} =____.$
Commented by mr W last updated on 23/Jun/19

$c o u l d b e a n y v a l u e i n (- \infty, \infty) e x c e p t z e r o .$
	Answered by MJS last updated on 23/Jun/19
	$3^y =6^z ⇒ z=((ln 3)/(ln 6))y 2^x =3^y ⇒ y=((ln 2)/(ln 3))x ⇒ z=((ln 2)/(ln 6))x (1/x)+(1/y)+(1/z)=((2ln 6)/(xln 2)) with x∈C\0$
	$3^{y} = 6^{z} \Rightarrow z = \frac{\ln 3}{\ln 6} y$ $2^{x} = 3^{y} \Rightarrow y = \frac{\ln 2}{\ln 3} x \Rightarrow z = \frac{\ln 2}{\ln 6} x$ $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{2 \ln 6}{x \ln 2} with x \in C ∖ 0$
	Answered by $@ty@m last updated on 23/Jun/19

	$2^{x} = 3^{y} = 6^{z} = k, s a y$ $2 = k^{\frac{1}{x}}$ $3 = k^{\frac{1}{y}}$ $6 = k^{\frac{1}{z}}$ $∵ 2 \times 3 = 6$ $∴ k^{\frac{1}{x} + \frac{1}{y}} = k^{\frac{1}{z}}$ $\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$ $P l . r e c h e c k q u e s t i o n .$
	Commented by Tony Lin last updated on 23/Jun/19

	$i f t h e q u e s t i o n i s 2^{x} = 3^{y} = 6^{- z}$ $t h e n \frac{1}{x} + \frac{1}{y} + \frac{1}{z} w i l l b e 0$
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