Question

(a×a) + (b×b) = 7ab , show that log (a+b)/3 = {log(a)+log(b)}/2

Answer 1

$\underline{ \large{\text{Answer}}} :$

$\text{Given that,} \\ \\ (\text{a} \times \text{a}) + ( \text{b} \times \text{b}) = \text{7ab} \\ \\ \to { \text{a}}^{2} + { \text{b}}^{2} = \text{7ab} \\ \\ \to { \text{a}}^{2} + { \text{b}}^{2} + \text{2ab} = \text{7ab + 2ab} \\ \\ \to { \text{(a + b)}}^{2} = \text{9ab} \\ \\ \to \text{a + b} = 3 { \text{(ab)}}^{ \frac{1}{2} } \\ \\ \to \frac{ \text{a + b}}{3} = { \text{ab}}^{ \frac{1}{2} }$

$\text{Taking (log) to both sides, we get -} \\ \\ \log( \frac{ \text{a + b}}{3} ) = \log { \text{(ab)}}^{ \frac{1}{2} } \\ \\ \to \boxed{\log (\frac{ \text{a + b}}{3} ) = \frac{1}{2} (\text{loga + logb) }}$

$\text{Hence, proved.} \\ \\ \bigstar \: \underline{ \large { \text{MarkAsBrainliest}}} \: \bigstar$