If 2^x = 3^y = 6^-z, show that: $\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0$

Given,

2^x = 3^y = 6^-z

Let 2^x = 3^y = 6^-z = k,

First term,

⇒ 2^x = k

⇒ x = 1

⇒ 2 = $k^{\dfrac{1}{x}}$

Second term,

⇒ 3^y = k

⇒ y = 1

⇒ 3 = $k^{\dfrac{1}{y}}$

Third term,

⇒ 6^-z = k

⇒ z = -1

⇒ 6 = $k^{-\dfrac{1}{z}}$

Factorizing 6, we get :

⇒ 6 = 2 × 3

We have $2 = k^{\dfrac{1}{x}}, 3 = k^{\dfrac{1}{y}}, 6 = k^{-\dfrac{1}{z}}$,

$\Rightarrow k^{-\dfrac{1}{z}} = k^{\dfrac{1}{x}} \times k^{\dfrac{1}{y}} \\[1em] \Rightarrow k^{-\dfrac{1}{z}} = k^{\dfrac{1}{x}} \times k^{\dfrac{1}{y}}\\[1em] \text{ Equating the exponents}, \\[1em] \Rightarrow -\dfrac{1}{z} = \dfrac{1}{x} + \dfrac{1}{y} \\[1em] \Rightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0.$

Hence proved, $\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0$.