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

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

$\therefore 2^x = k \text{ or } 2 = k^{\dfrac{1}{x}}$ ……(i)

$\therefore 3^y = k \text{ or } 3 = k^{\dfrac{1}{y}}$ ……(ii)

$\therefore 6^{-z} = k \text{ or } 6 = k^{-\dfrac{1}{z}}$ ……(iii)

We know that,

2 × 3 = 6

Substituting value of 2, 3 and 6 from (i), (ii) and (iii) in above equation we get,

$\Rightarrow k^{\dfrac{1}{x}} \times k^{\dfrac{1}{y}} = k^{-\dfrac{1}{z}} \\[1em] \Rightarrow k^{\dfrac{1}{x} + \dfrac{1}{y}} = k^{-\dfrac{1}{z}} \\[1em] \therefore \dfrac{1}{x} + \dfrac{1}{y} = -\dfrac{1}{z} \\[1em] \Rightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0.$

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