Question

If $a^2-12ab+4b^2=0$, prove that
$\log (a+2b)=\frac{1}{2}(\log a+\log b)+2\log 2$

Answer 1

We have,

$a^2-12ab+4b^2=0$

$(a+2b{)}^2-16ab=0$

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

Taking log on both sides inthe above equation,

$2log(a+2b)=log\left(16ab\right)$

$log(a+2b)=\frac{1}{2}log\left(16ab\right)$

$log(a+2b)=\frac{1}{2}log\left(16\right)+\frac{1}{2}(loga+logb)$

$log(a+2b)=log\left(4\right)+\frac{1}{2}(loga+logb)$

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

Hence it proved.