Logarithmic Prove

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))

 

Thank you so much. You are the best.