Question

If $\log \left(\frac{x-y}{4}\right)=\log \sqrt{x}+\log \sqrt{y}$, show that $(x+y{)}^2=20xy$

Answer 1

We have,

$\log \left(\frac{x-y}{4}\right)=\log \sqrt{x}+\log \sqrt{y}$

$\Rightarrow \log \left(\frac{x-y}{4}\right)=\log {\left(x\right)}^{\frac{1}{2}}+\log {\left(y\right)}^{\frac{1}{2}}$

$\Rightarrow \log \left(\frac{x-y}{4}\right)=\frac{1}{2}\log x+\frac{1}{2}\log y$

$\Rightarrow \log \left(\frac{x-y}{4}\right)=\frac{1}{2}(\log x+\log y)$

$\Rightarrow \log \left(\frac{x-y}{4}\right)=\frac{1}{2}\log xy\therefore \log x+\log y=\log xy$

$\Rightarrow 2\log \left(\frac{x-y}{4}\right)=\log xy$

$\Rightarrow \log {\left(\frac{x-y}{4}\right)}^2=\log xy$

On comparing both side and we get,

$\Rightarrow {\left(\frac{x-y}{4}\right)}^2=xy$

$\Rightarrow {(x-y)}^2=16xy$

$\Rightarrow {(x+y)}^2-4xy=16xy$

$\Rightarrow {(x+y)}^2=16xy+4xy$

$\Rightarrow {(x+y)}^2=20xy$

Hence, proved.