Question

help me solve this please​

Answer 1

Heya!

Given,

a²+b² = 7ab

by adding "2ab" on both sides,

a²+b²+2ab = 7ab+2ab

(a+b)² = 9ab

By taking "log",

log (a+b)² = log 9ab

2 log(a+b) = log 3²ab

2 log(a+b) = log 3² + log a + log b

2 log(a+b) = 2 log 3 + log a + log b

2 log(a+b) - 2 log 3 = log a + log b

2 [log(a+b) - log 3] = log a + log b

log(a+b) - log 3 = (log a + log b) /2

log(a+b/3) = 1/2 log a + 1/2 log b

Answer 2

Correct Question :-

If a² + b² = 7ab, show that

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

Solution :-

a² + b² = 7ab

Adding 2ab on both sides

a² + b² + 2ab = 7ab + 2ab

(a + b)² = 9ab

[ Because (a + b)² = a² + b² + 2ab]

Taking log on both sides

$\\ \\ \sf log(a + b)^2 = log9ab$

$\\ \\ \sf \longrightarrow log(a + b)^2 = log9ab \\ \\ \\ \sf \longrightarrow log(a + b)^2 = log9 + loga + logb \\ \\ \\ \boxed{ \bf \because log(ab) = loga + logb} \\ \\ \\ \sf \longrightarrow log(a + b)^2 = log3^2 + loga + logb \\ \\ \\ \sf \longrightarrow 2log(a + b) = 2log3 + loga + logb \\ \\ \\ \bf \underline{Transpose \: 2log3 \: to \: LHS} \\ \\ \\ \sf \longrightarrow 2log(a + b) - 2log3 = loga + logb \\ \\ \\ \sf \longrightarrow 2 \bigg[log(a + b) - log3 \bigg] = loga + logb$

$\\ \\ \sf \longrightarrow log(a + b) - log3 = \dfrac{1}{2} \bigg( loga + logb \bigg) \\ \\ \\ \sf \longrightarrow log \bigg( \dfrac{a + b}{3} \bigg) = \dfrac{1}{2}loga + \dfrac{1}{2}logb \\ \\ \\ \boxed{ \bf \because loga - logb = log \bigg( \dfrac{a}{b} \bigg)}$

Hence shown