If $\dfrac{1}{\text{ log }$a x} + \dfrac{1}{\text{ log }b x} = \dfrac{2}{\text{ log }_c x}, prove that c² = ab.

Given,

$\Rightarrow \dfrac{1}{\text{ log }$a x} + \dfrac{1}{\text{ log }b x} = \dfrac{2}{\text{ log }_c x}\\[1em] \Rightarrow \dfrac{1}{\dfrac{\text{log }x}{\text{log }a}} + \dfrac{1}{\dfrac{\text{log } x}{\text{log } b}} = \dfrac{2}{\dfrac{\text{log }x}{\text{log }c}} \\[1em] \Rightarrow \dfrac{\text{ log } a}{\text{ log } x} + \dfrac{\text{ log } b}{\text{ log } x} = \dfrac{2\text{ log } c}{\text{ log } x}\\[1em] \Rightarrow \dfrac{\text{ log } a + \text{ log } b}{\text{ log } x} = \dfrac{2\text{ log } c}{\text{ log } x}\\[1em] \Rightarrow \dfrac{\text{ log } (a \times b)}{\text{ log } x} = \dfrac{2\text{ log } c}{\text{ log } x}\\[1em] \Rightarrow \dfrac{\text{ log } ab}{\text{ log } x} = \dfrac{2\text{ log } c}{\text{ log } x}\\[1em] \Rightarrow \text{ log } ab = 2\text{ log } c\\[1em] \Rightarrow \text{ log } ab = \text{ log } c^2\\[1em] \Rightarrow ab = c^2.

Hence, proved that c² = ab.