Question

If $\log \left(\frac{x-y}{4}\right)=\log \sqrt{x}+\log \sqrt{y},$. Find $m$ if ${(x+y)}^2=mxy$

Answer 1

$\frac{x-y}{4}=\sqrt{x}\sqrt{y}$

$\frac{\sqrt{x}}{\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}}=4$

Now, let

$\sqrt{\frac{x}{y}}=zz-\frac{1}{z}=4$

$z^2-1=4z$

$z^2-4z-1=0$

Now,zeroes of this equation are

$\sqrt{\frac{x}{y}}and\sqrt{\frac{y}{x}}$

$So,\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=4$

Now,

${(x+y)}^2=mxy$

${\left(\frac{x+y}{\sqrt{x}\sqrt{y}}\right)}^2=m$

${(\sqrt{\frac{y}{x}}+\sqrt{\frac{x}{y}})}^2=m$

$y^2=16$

$m=16$