Question

If $\log \left(\frac{a+b}{4}\right)=\frac{1}{2}[\log a+\log b]$ , show that $(a-b{)}^2=12ab$

Answer 1

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

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

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

$\Rightarrow \frac{a+b}{4}=\sqrt{ab}$

$a+b=4\sqrt{ab}$

Squaring both sides,

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

Now,

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

$=16ab-4ab$

$=12ab$