Question

If a^2+ab+b^2=0 ,Prove that Log(a+b)=1/2 (loga+logb)

Answer 1

I'am just checking da answer .....

Answer 2

OH sure, why don't you add your work then?

Answer 3

\[(a+b)^2=a^2+2ab+b^2=ab\] is a start

Answer 4

should work neatly from there

Answer 5

\[ a+b = \sqrt{ab} \implies \log (a+b) =\frac 12 \log (ab) = \frac 12 (\log a+\log b) \]

Answer 6

here is it :
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a^2+ab+b^2=0 , (add ab and subtract ab)
a^2+2ab+b^2=ab ,(a+b)^2=ab ..Then take log the da both sides
log(a+b)^2=log(ab) ,2log(a+b)=log(ab) (divide by 2)
log(a+b)=1/2(loga+logb)
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Answer 7

am i doing right ?

Answer 8

yes

Answer 9

good ty