Solve for x:

log₂ x + log₈ x + log₃₂ x = $\dfrac{23}{15}$

Given,

$\Rightarrow \text{log}$2 \space x + \text{log}8 \space x + \text{log}{32} \space x = \dfrac{23}{15} \\[1em] \Rightarrow \dfrac{1}{\text{log}x \space 2} + \dfrac{1}{\text{log}x \space 8} + \dfrac{1}{\text{log}x \space 32} = \dfrac{23}{15} \\[1em] \Rightarrow \dfrac{1}{\text{log}x \space 2} + \dfrac{1}{\text{log}x \space 2^3} + \dfrac{1}{\text{log}x \space 2^5} = \dfrac{23}{15} \\[1em] \Rightarrow \dfrac{1}{\text{log}x \space 2} + \dfrac{1}{3\text{log}x \space 2} + \dfrac{1}{5\text{log}x \space 2} = \dfrac{23}{15} \\[1em] \Rightarrow \dfrac{1}{\text{log}x \space 2}\Big[1 + \dfrac{1}{3} + \dfrac{1}{5}\Big] = \dfrac{23}{15} \\[1em] \Rightarrow \text{log}2 \space x\Big[\dfrac{15 + 5 + 3}{15}\Big] = \dfrac{23}{15} \\[1em] \Rightarrow \text{log}2 \space x \times \dfrac{23}{15} = \dfrac{23}{15} \\[1em] \Rightarrow \text{log}2 \space x = \dfrac{23}{15} \times \dfrac{15}{23} \\[1em] \Rightarrow \text{log}2 \space x = 1 \\[1em] \Rightarrow \text{log}2 \space x = \text{log}_2 \space 2 \\[1em] \Rightarrow x = 2.

Hence, x = 2.