If 2^x = 3^y = 12^z, show that: $\dfrac{1}{z} = \dfrac{1}{y} + \dfrac{2}{x}$

Given,

2^x = 3^y = 12^z

Let 2^x = 3^y = 12^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,

⇒ 12^z = k

⇒ z = 1

⇒ 12 = $k^{\dfrac{1}{z}}$

Factorizing 12, we get :

⇒ 12 = 2² × 3

We have $2 = k^{\dfrac{1}{x}}, 3 = k^{\dfrac{1}{y}}, 12 = k^{\dfrac{1}{z}}$,

$\Rightarrow k^{\dfrac{1}{z}} = \Big(k^{\dfrac{1}{x}}\Big)^2 + k^{\dfrac{1}{y}} \\[1em] \Rightarrow k^{\dfrac{1}{z}} = \Big(k^{\dfrac{2}{x}}\Big) + k^{\dfrac{1}{y}} \\[1em] \text{ Equating the exponents}, \\[1em] \Rightarrow \dfrac{1}{z} = \dfrac{2}{x} + \dfrac{1}{y}$

Hence proved, $\dfrac{1}{z} = \dfrac{1}{y} + \dfrac{2}{x}$.