If log $\dfrac{a - b}{2} = \dfrac{1}{2}$ (log a + log b), show that :

a² + b² = 6ab

Given,

$\Rightarrow \text{log } \dfrac{a - b}{2} = \dfrac{1}{2} \text{(log a + log b)} \\[1em] \Rightarrow \text{log } \dfrac{a - b}{2} = \dfrac{1}{2} (\text{log ab}) \\[1em] \Rightarrow \text{log } \dfrac{a - b}{2} = \text{log (ab)} ^{\dfrac{1}{2}} \\[1em] \Rightarrow \dfrac{a - b}{2} = (ab)^{\dfrac{1}{2}}$

Squaring both sides, we get :

$\Rightarrow \Big(\dfrac{a - b}{2}\Big)^2 = (ab)^{\dfrac{1}{2} \times 2} \\[1em] \Rightarrow \dfrac{a^2 + b^2 - 2ab}{4} = ab \\[1em] \Rightarrow a^2 + b^2 - 2ab = 4ab \\[1em] \Rightarrow a^2 + b^2 = 4ab + 2ab \\[1em] \Rightarrow a^2 + b^2 = 6ab.$

Hence, proved that a² + b² = 6ab.