If a^2+b^2=7ab prove log 1/3(a+b)=1/2(log a+log b) With proper steps Pleeeease?

Answer 1

if a^2 + b^2 = 7ab => a^2 + 2ab + b^2 = 9ab, or (a+b)^2 = 9ab.

so log (a+b)^2 = log(9ab) = log 9 + log a + log b,
then 2 log(a+b) = 2 log 3 + log a + log b,
=> log(a+b) - log 3 = (log a + log b)/2
=> log(a+b)/3 = (log a + log b)/2.

Answer 2

a^2 + b^2 = 7ab means a^2 + 2ab + b^2 = 9ab, or (a+b)^2 = 9ab.

Therefore log (a+b)^2 = log 9 + log a + log b,
or 2 log(a+b) = 2 log 3 + log a + log b,
meaning log(a+b) - log 3 = (log a + log b)/2
so log(a+b)/3 = (log a + log b)/2.

Answer 3

>a^2+b^2+2ab=9ab
>(a+b)^2=9ab
>2log(a+b)=log(9ab)
>log(a+b)=1/2(log9)+{1/2[(log a)+ (log b)]}
>log(a+b)=(1/2)*2*log3 +{1/2[(log a)+ (log b)]}
>log(a+b)=log3+{1/2[(log a)+ (log b)]}
>log(a+b)-log3={1/2[(log a)+ (log b)]}
>log{(a+b)/3} ={1/2[(log a)+ (log b)]}

hence proved ...u missed a log on the left hand side ..

Answer 4

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