If-a-3-b-3-0-prove-that-log-a-b-1-2-log-a-log-b-log-3-given-a-b-0-

Question Number 40872 by scientist last updated on 28/Jul/18

I f a^{3} + b^{3} = 0, p r o v e t h a t \log (a + b) = \frac{1}{2} (\log a + \log b + \log 3)

[g i v e n a + b \neq 0]

Commented by MrW3 last updated on 29/Jul/18

I {d o n}^{'} t t h i n k t h e r e a r e s u c h r e a l v a l u e s

f o r a a n d b .

s u c h t h a t \log (a + b) = \frac{1}{2} (\log a + \log b + \log 3)

i s d e f i n e d, w e h a v e a > 0, b > 0, t h e n

w e g e t a^{3} + b^{3} > 0, b u t a^{3} + b^{3} = 0 .

Commented by maxmathsup by imad last updated on 29/Jul/18

y o u a r e r i g h t s i r i t s a c o m p l e x l o g a r i t h m e b e c a u s e a^{2} - a b + b^{2} > 0

Answered by tanmay.chaudhury50@gmail.com last updated on 29/Jul/18

a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2}) = 0

g i v e n a + b n o t e q u a l s t o 0

s o a^{2} - a b + b^{2} = 0

{(a + b)}^{2} - 2 a b - a b = 0

a + b = \sqrt{3 a b}

l n (a + b) = \frac{1}{2} (l n a + l n b + l n 3)