Question

If $log2+\frac{1}{2}loga+\frac{1}{2}logb=log(a+b)$, then which of the following holds true?

• A. a = b
• B. a = -b
• C. a = 2, b = 0
• D. a = 10, b = 1

Answer 1

The correct option is A

a = b

$log2+\frac{1}{2}loga+\frac{1}{2}logb=log(a+b)$

$log2\sqrt{ab}=log(a+b)$

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

Squaring on both the sides

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

$\to (a+b{)}^2=0$

$\to a=b$