If 2^x = 3^y = 12^z, prove that x = $\dfrac{2yz}{y - z}.$

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

$\therefore 12^{z} = k \text{ or } 12 = k^{\dfrac{1}{z}}$ ……(iii)

We know that,

2² × 3 = 12

Substituting value of 2, 3 and 12 from (i), (ii) and (iii) in above equation we get,

$\Rightarrow (k^{\dfrac{1}{x}})^2 \times k^{\dfrac{1}{y}} = k^{\dfrac{1}{z}} \\[1em] \Rightarrow k^{\dfrac{2}{x}} \times k^{\dfrac{1}{y}} = k^{\dfrac{1}{z}} \\[1em] \Rightarrow k^{\dfrac{2}{x} + \dfrac{1}{y}} = k^{\dfrac{1}{z}} \\[1em] \therefore \dfrac{2}{x} + \dfrac{1}{y} = \dfrac{1}{z} \\[1em] \Rightarrow \dfrac{2}{x} = \dfrac{1}{z} - \dfrac{1}{y} \\[1em] \Rightarrow \dfrac{2}{x} = \dfrac{y - z}{yz} \\[1em] \Rightarrow \dfrac{x}{2} = \dfrac{yz}{y - z} \\[1em] \Rightarrow x = \dfrac{2yz}{y - z}.$

Hence, proved that $x = \dfrac{2yz}{y - z}.$