If a^2 + b^2 =7ab , prove log (a+b/3)=1/2 (loga+logb)

Answer 1

Please write the problem less ambiguously: If

a² + b² = 7ab

Prove

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

The a+b/3 threw me off for a bit, so I almost didn't respond.

The first two things to consider is simplifying (log a + log b) and taking care of the (1/2) by multiplying both sides by 2.

Remember two important log rules

1. log xy = log x + log y
2. n log x = log xⁿ

Notice that the given includes a² + b², as well as terms with ab, so it is also worth remembering that

1. a² + 2ab + b² = (a + b)²

The "creative problem solving" part is to add 2ab to both sides, then things should fall in to place.

Answer 2

a^2+b^2=7ab

add 2ab to both sides

a^2+b^2+2ab=9ab

(a+b)^2=9ab

prove log(a+b)/3=(1/2)(loga+logb)

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

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

(a+b)/3=(ab)^(1/2)

square both sides

(a+b)^2/9=ab

(a+b)^2=9ab

9ab=9ab from above