- Joined
- Jun 27, 2021
- Messages
- 5,386
- Reaction score
- 422
if a^2+b^2=7ab
show that
log((1/3)(a+b))=(1/2)(log(a)+log(b))
Remember that:
(a+b)^2=a^2+2ab+b^2
and
log(a)+log(b)=log(ab)
so,
if a^2+b^2=7ab
adding 2ab on both sides :
a^2+b^2+2ab=7ab+2ab
(a+b)^2=9ab
a+b=sqrt(9ab)
a+b=3sqrt(ab)
Multiply by 1/3 on both sides :
(1/3)(a+b)=(1/3)3sqrt(ab)
(1/3)(a+b)=sqrt(ab)
(1/3)(a+b)=(ab)^(1/2)
take log on both sides :
log((1/3)(a+b))=log((ab)^(1/2))
log((1/3)(a+b))=(1/2)log(ab)
since (1/2)log(ab)=(1/2)(log(a)+log(b))
so,
log((1/3)(a+b))=(1/2)(log(a)+log(b))