Given that log_a x = $\dfrac{1}{α}$, log_b x = $\dfrac{1}{β}$, log_c x = $\dfrac{1}{γ}$, find log_abc x.

Given,

$\Rightarrow \dfrac{1}{α} = \text{log}$a\space x = \dfrac{\text{log x}}{\text{log a}} \\[1em] \Rightarrow \text{log a} = α\text{log x} \\[1em] \\[1em] \Rightarrow \dfrac{1}{β} = \text{log}b\space x = \dfrac{\text{log x}}{\text{log b}} \\[1em] \Rightarrow \text{log b} = β\text{log x} \\[1em] \\[1em] \Rightarrow \dfrac{1}{γ} = \text{log}_c\space x = \dfrac{\text{log x}}{\text{log c}} \\[1em] \Rightarrow \text{log c} = γ\text{log x} \\[1em] \\[1em]

Solving log_abc x we get,

$\Rightarrow \text{log}_\text{abc}\space x = \dfrac{\text{log x}}{\text{log abc}} \\[1em] = \dfrac{\text{log x}}{\text{log a + log b + log c}} \\[1em] = \dfrac{\text{log x}}{\text{αlog x + βlog x + γlog x}} \\[1em] = \dfrac{\text{log x}}{\text{log x(α + β + γ)}} \\[1em] = \dfrac{1}{(\text{α} + \text{β} + \text{γ})}.$

Hence, log_abc x = $\dfrac{1}{(\text{α} + \text{β} + \text{γ})}$.