The pair of equations $3^{x+y}=81$ and $81^{x-y}=3$ has:
$\textbf{(A)}\ \text{no common solution} \qquad \\ \textbf{(B)}\ \text{the solution} \text{ } x=2, y=2\qquad \\ \textbf{(C)}\ \text{the solution} \text{ } x=2\frac{1}{2}, y=1\frac{1}{2} \qquad \\ \textbf{(D)}\text{ a common solution in positive and negative integers} \qquad \\ \textbf{(E)}\ \text{none of these}$
Rewrite the equations so both sides have a common base.
\[3^{x+y}=3^4\]
\[3^{4(x-y)}=3^1\]
Taking the logarithm base 3, we get a linear [system of equations](https://artofproblemsolving.com/wiki/index.php/System_of_equations).
\[x+y=4\]
\[4x-4y=1\]
Solve the system to get $x=\frac{17}{8}$ and $y=\frac{15}{8}$.  The answer is $\boxed{\textbf{(E)}}$.