If a^3 +b^3 =0, prove that log (a+b)=(1/2)(log a +log b +log 3) [given a+b≠0]


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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

	$a^{2} + b^{2} = a b$ $a^{2} + b^{2} + 2 a b = a b + 2 a b [a d d 2 a b b o t h s i d e]$ $∴ {(a + b)}^{2} = 3 a b$ $I s i t c l e a r ?$
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