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

(Solution to Problem 18)
 
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== Problem 18==
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==Problem==
  
 
The pair of equations <math>3^{x+y}=81</math> and <math>81^{x-y}=3</math> has:
 
The pair of equations <math>3^{x+y}=81</math> and <math>81^{x-y}=3</math> has:

Revision as of 20:23, 9 May 2018

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 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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