Let $x$ and $y$ be positive real numbers such that ${\log }_5(x+y)+{\log }_5(x-y)=3$ and ${\log }_{2}y-{\log }_{2}x=1-{\log }_{2}3$. Then $xy$ equals

Correct Option: 3 (which corresponds to 100)

Let's simplify the first equation:
${\log }_5(x+y)+{\log }_5(x-y)=3$

Using the property of logarithms $\log \left(a\right)+\log \left(b\right)=\log \left(ab\right)$:
${\log }_5\left(\right(x+y\left)\right(x-y\left)\right)=3$
${\log }_5(x^2-y^2)=3$
$x^2-y^2=5^3=125$ --- (Equation 1)

Now let's simplify the second equation:
${\log }_{2}y-{\log }_{2}x=1-{\log }_{2}3$

Using the property $\log \left(a\right)-\log \left(b\right)=\log \left(\frac{a}{b}\right)$:
${\log }_2\left(\frac{y}{x}\right)={\log }_{2}2-{\log }_{2}3$
${\log }_2\left(\frac{y}{x}\right)={\log }_2\left(\frac{2}{3}\right)$
$\frac{y}{x}=\frac{2}{3}$
$y=\frac{2x}{3}$ --- (Equation 2)

Substitute Equation 2 into Equation 1:
$x^2-{\left(\frac{2x}{3}\right)}^2=125$
$x^2-\frac{4x^2,}{9}=125$
$\frac{5x^2}{9}=125$
$x^2=\frac{125\times 9}{5}=25\times 9=225$

Since $x$ must be a positive real number, we have:
$x=15$

Now find $y$ using Equation 2:
$y=\frac{2\left(15\right)}{3}=10$

Therefore, $xy$ is:
$xy=15\times 10=150$

Let's double check the option values. The calculation shows $xy=150$, which corresponds to Option 4.