Question

If $a^{2} + 4b^{2} = 12ab,$ then $\log(a + 2b)$ is

• A. $\frac{1}{2}\lbrack\log a + \log b - \log 2\rbrack$
• B. $\log\frac{a}{2} + \log\frac{b}{2} + \log 2$
• C. $\frac{1}{2}\lbrack\log a + \log b + 4\log 2\rbrack$
• D. $\frac{1}{2}\lbrack\log a - \log b + 4\log 2\rbrack$

Answer 1

Answer: (C)

$\frac{1}{2}\lbrack\log a + \log b + 4\log 2\rbrack$

Answer 2

$a^{2} + 4b^{2} = 12ab$ $\Rightarrow$ $a^{2} + 4b^{2} + 4ab = 16ab$

$\Rightarrow (a + 2b)^{2} = 16ab$

$\Rightarrow 2\log(a + 2b) = \log 16 + \log a + \log b$

$\therefore$ $\log(a + 2b) = \frac{1}{2}\lbrack\log a + \log b + 4\log 2\rbrack$