# Difference between revisions of "1960 AHSME Problems/Problem 18"

## Problem

The pair of equations $3^{x+y}=81$ and $81^{x-y}=3$ has: $\textbf{(A)}\ \text{no common solution} \qquad \\ \textbf{(B)}\ \text{the solution} \text{ } x=2, y=2\qquad \\ \textbf{(C)}\ \text{the solution} \text{ } x=2\frac{1}{2}, y=1\frac{1}{2} \qquad \\ \textbf{(D)}\text{ a common solution in positive and negative integers} \qquad \\ \textbf{(E)}\ \text{none of these}$

## Solution

Rewrite the equations so both sides have a common base. $$3^{x+y}=3^4$$ $$3^{4(x-y)}=3^1$$ This results in a linear 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 $\boxed{\textbf{(E)}}$.

## See Also

 1960 AHSC (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 19 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 All AHSME Problems and Solutions
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