If a² + b² = 23ab, show that :

log $\dfrac{a + b}{5} = \dfrac{1}{2}$ (log a + log b).

Given,

⇒ a² + b² = 23ab

⇒ a² + b² + 2ab = 23ab + 2ab

⇒ (a + b)² = 25ab

⇒ $\dfrac{(a + b)^2}{25}$ = ab

⇒ $\Big(\dfrac{a + b}{5}\Big)^2$ = ab

Taking log on both sides, we get :

⇒ log $\Big(\dfrac{a + b}{5}\Big)^2$ = log ab

⇒ 2 log $\dfrac{a + b}{5}$ = log a + log b

⇒ log $\dfrac{a + b}{5}$ = $\dfrac{1}{2}$ (log a + log b).

Hence, proved that log $\dfrac{a + b}{5}$ = $\dfrac{1}{2}$ (log a + log b).