20. The real numbers $x$ and $y$ satisfy the equations $4^{y}=\frac{1}{8(\sqrt{2})^{x+2}}$ and $9^{x} \times 3^{y}=3 \sqrt{3}$. What is the value of $5^{x+y}$ ?
A $5 \sqrt{5}$
B 5
C $\sqrt{5}$
D $\frac{1}{5}$
E $\frac{1}{\sqrt{5}}$
Solution
E
We have $4^{y}=\left(2^{2}\right)^{y}=2^{2 y}$, and $\frac{1}{8(\sqrt{2})^{x+2}}=\frac{1}{2^{3}\left(2^{\frac{1}{2}}\right)^{x+2}}=2^{-\left(3+\frac{1}{2}(x+2)\right)}$.
Therefore from the equation $4^{y}=\frac{1}{8(\sqrt{2})^{x+2}}$ we deduce that $2 y=-\left(3+\frac{1}{2}(x+2)\right)$.
Therefore
$$
y=-\frac{1}{4} x-2 .
$$

Also, $9^{x} \times 3^{y}=\left(3^{2}\right)^{x} \times 3^{y}=3^{2 x+y}$ and $3 \sqrt{3}=\left(3^{1}\right)\left(3^{\frac{1}{2}}\right)=3^{1+\frac{1}{2}}=3^{\frac{3}{2}}$.
Therefore, from the equation $9^{x} \times 3^{y}=3 \sqrt{3}$ we deduce that $2 x+y=\frac{3}{2}$.
Therefore
$$
y=-2 x+\frac{3}{2} \text {. }
$$

From equations (1) and (2)
$$
-\frac{1}{4} x-2=-2 x+\frac{3}{2} .
$$

This last equation may be rearranged to give
$$
\frac{7}{4} x=\frac{7}{2} .
$$

Hence $x=2$.
Therefore, from equation (1), $y=-\frac{5}{2}$.
It follows that $x+y=2+\left(-\frac{5}{2}\right)=2-\frac{5}{2}=-\frac{1}{2}$.
Therefore $5^{x+y}=5^{-\frac{1}{2}}=\frac{1}{\sqrt{5}}$.
For INVESTIGATION
20.1 The real numbers $x$ and $y$ satisfy the equations
$$
(\sqrt{3})^{x} \times 3^{y}=\sqrt[5]{27}
$$
and
$$
8^{x} \times(\sqrt{2})^{y}=\sqrt[3]{4}
$$

Find the value of $8^{x+y}$.