If 2^x = 4^y = 8^z and $\dfrac{1}{2x} + \dfrac{1}{4y} + \dfrac{1}{8z}$ = 4, find the value of x.

Given,

⇒ 2^x = 4^y = 8^z

⇒ 2^x = (2²)^y = (2³)^z

⇒ 2^x = 2^2y = 2^3z

⇒ x = 2y = 3z

⇒ x = 2y and x = 3z

$\Rightarrow y = \dfrac{x}{2} \text{ and } z = \dfrac{x}{3}$.

Substituting value of y and z in $\dfrac{1}{2x} + \dfrac{1}{4y} + \dfrac{1}{8z} = 4$, we get :

$\Rightarrow \dfrac{1}{2x} + \dfrac{1}{4 \times \dfrac{x}{2}} + \dfrac{1}{8 \times \dfrac{x}{3}} = 4 \\[1em] \Rightarrow \dfrac{1}{2x} + \dfrac{1}{2x} + \dfrac{3}{8x} = 4 \\[1em] \Rightarrow \dfrac{2}{2x} + \dfrac{3}{8x} = 4 \\[1em] \Rightarrow \dfrac{8 + 3}{8x} = 4 \\[1em] \Rightarrow \dfrac{11}{8x} = 4 \\[1em] \Rightarrow x = \dfrac{11}{8 \times 4} = \dfrac{11}{32}.$

Hence, x = $\dfrac{11}{32}$.