Question

If $a^2+4b^2=12ab$, then value of $\log (a+2b)$ is equal to

• A. $\frac{1}{2}\log \left(\frac{ab}{2}\right)$
• B. $\frac{1}{2}\log a+\frac{1}{2}\log b+2\log 2$
• C. $\frac{1}{2}\log \left(\frac{16a}{b}\right)$
• D. $\frac{1}{2}\log \left(\frac{ab}{16}\right)$

Answer 1

The correct option is B $\frac{1}{2}\log a+\frac{1}{2}\log b+2\log 2$

$a^2+4b^2=12ab$
$\Rightarrow a^2+4b^2+4ab=16ab$
$\Rightarrow {(a+2b)}^2=16ab$
Taking log both sides,
$\Rightarrow \log {(a+2b)}^2=\log 16ab$
$\Rightarrow 2\log (a+2b)=\log a+\log b+4\log 2[\because \log a^m=m\log a\text{\&}\log (abc)=\log a+\log b+\log c]$
$\Rightarrow \log (a+2b)=\frac{1}{2}\log a+\frac{1}{2}\log b+2\log 2$
Ans: B