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 40873 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]

Answered by math1967 last updated on 29/Jul/18

a^{3} + b^{3} = 0

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

a^{2} - a b + b^{2} = 0 [∵ (a + b) \neq 0]

a^{2} + b^{2} = a b

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

2 l o g (a + b) = l o g 3 a b

∴ l o g (a + b) = \frac{1}{2} (l o g a + l o g b + l o g 3)

Commented by FaHmYaSmA last updated on 27/Oct/18

how it can be 3ab ? where '3' come from ?

Commented by math1967 last updated on 27/Oct/18

a^{2} + b^{2} = a b

a^{2} + b^{2} + 2 a b = a b + 2 a b [a d d 2 a b b o t h s i d e]

∴ {(a + b)}^{2} = 3 a b

I s i t c l e a r ?