If a, b are positive real numbers, a > b and a² + b² = 27ab, prove that

$\text{log} \space\Big(\dfrac{a - b}{5}\Big) = \dfrac{1}{2}(\text{log} \space a + \text{log} \space b)$

Given,

a² + b² = 27ab

⇒ a² + b² = 2ab + 25ab

⇒ a² + b² - 2ab = 25ab

$\Rightarrow ab = \dfrac{a^2 + b^2 - 2ab}{25} = \Big(\dfrac{a - b}{5}\Big)^2$

Taking log on both sides:

$\Rightarrow \text{log} \space ab = \text{log} \space \Big(\dfrac{a - b}{5}\Big)^2 \\[1em] \Rightarrow \text{log} \space a + \text{log} \space b = 2 \text{log} \space \Big(\dfrac{a - b}{5}\Big) \\[1em] \Rightarrow \text{log} \space \Big(\dfrac{a - b}{5}\Big) = \dfrac{1}{2}(\text{log} \space a + \text{log} \space b).$

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