Question

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

• A. $\frac{1}{2}[loga+logb-log2]$
• B. $log\frac{a}{2}+log\frac{b}{2}+log2$
• C. $\frac{1}{2}[loga+logb+4log2]$
• D. $\frac{1}{2}[loga-logb+4log2]$

Answer 1

The correct option is C $\frac{1}{2}[loga+logb+4log2]$

$a^2+4b^2=12ab\Rightarrow a^2+4b^2+4ab=16ab\Rightarrow (a+2b{)}^2=16ab$
$\Rightarrow 2log(a+2b)=log16+loga+logb$
$\therefore log(a+2b)=\frac{1}{2}[loga+logb+4log2]$