If log2+12loga+12logb=log(a+b), then

log2+12loga+12logb=log(a+b)log2ab=log(a+b)2ab=a+bSquaring on both sides4ab=(a+b)2⇒(a−b)2=0a=b

If log2+12loga+12logb=log(a+b), then

log2+12loga+12logb=log(a+b)log2ab=log(a+b)2ab=a+bSquaring on both sides4ab=(a+b)2⇒(a−b)2=0a=b

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If $\log 2 + \frac{1}{2} \log a + \frac{1}{2} \log b = \log (a + b)$ , then

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$a = b$

$a = - b$

$a = 2, b = 0$

$a = 10, b = 2$

answer is A.

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Detailed Solution

$\log 2 + \frac{1}{2} \log a + \frac{1}{2} \log b = \log (a + b)$

$\log 2 \sqrt{a b} = \log (a + b)$

$2 \sqrt{a b} = a + b$

Squaring on both sides

$4 a b = {(a + b)}^{2}$

$\Rightarrow {(a - b)}^{2} = 0$

$a = b$

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