Question

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

• A. $\frac{\text{1}}{\text{2}}\left[\text{log}\text{a}\text{+}\text{log}\text{b}\text{-}\text{log2}\right]$
• B. $\text{log}\frac{\text{a}}{\text{2}}\text{+}\text{log}\frac{\text{b}}{\text{2}}\text{+}\text{log2}$
• C. $\frac{\text{1}}{\text{2}}\left|\text{log}\text{a}\text{+}\text{log}\text{b + 4}\text{log}\text{2}\right|$
• D. $\frac{\text{1}}{\text{2}}\left|\text{log}\text{a}-\text{log}\text{b + 4}\text{log}\text{2}\right|$

Answer 1

Answer: (C)

$\frac{\text{1}}{\text{2}}\left|\text{log}\text{a}\text{+}\text{log}\text{b + 4}\text{log}\text{2}\right|$

We have,

$a^2+4b^2=12ab$

$a^2+4b^2+4ab=16ab$

$a^2+(2b{)}^2+4ab=16ab$

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

On taking logarithm both sides, we get

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

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

$2\log (a+2b)=\log 2^4+\log a+\log b$

$2\log (a+2b)=4\log 2+\log a+\log b$

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

Hence, this is the answer.