If a² + b² = 7ab, prove that log ((a+b)/(3) ) = (1)/(2)(log a + log b) — JEE Mathematics

Add $2ab$ to both sides of the given equation to complete the square on the left:

$$a^2 + b^2 + 2ab = 7ab + 2ab$$

$$(a + b)^2 = 9ab$$

Divide both sides by 9:

$$\frac{(a + b)^2}{9} = ab \implies \left(\frac{a + b}{3}\right)^2 = ab$$

Take the natural logarithm (or base 10) on both sides:

$$\log\left(\frac{a + b}{3}\right)^2 = \log(ab)$$

Using logarithmic properties ($\log(x^2) = 2\log x$ and $\log(xy) = \log x + \log y$):

$$2\log\left(\frac{a + b}{3}\right) = \log a + \log b$$

Divide by 2:

$$\log\left(\frac{a + b}{3}\right) = \frac{1}{2}(\log a + \log b)$$

(Hence Proved)