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

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<cmath>3^{x+y}=3^4</cmath>
 
<cmath>3^{x+y}=3^4</cmath>
 
<cmath>3^{4(x-y)}=3^1</cmath>
 
<cmath>3^{4(x-y)}=3^1</cmath>
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This results in a linear system of equations.
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Taking the logarithm base 3, we get a linear [[system of equations]].
 
<cmath>x+y=4</cmath>
 
<cmath>x+y=4</cmath>
 
<cmath>4x-4y=1</cmath>
 
<cmath>4x-4y=1</cmath>
 
Solve the system to get <math>x=\frac{17}{8}</math> and <math>y=\frac{15}{8}</math>.  The answer is <math>\boxed{\textbf{(E)}}</math>.
 
Solve the system to get <math>x=\frac{17}{8}</math> and <math>y=\frac{15}{8}</math>.  The answer is <math>\boxed{\textbf{(E)}}</math>.
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==See Also==
 
==See Also==
 
{{AHSME 40p box|year=1960|num-b=17|num-a=19}}
 
{{AHSME 40p box|year=1960|num-b=17|num-a=19}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 21:14, 13 January 2023

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. \[3^{x+y}=3^4\] \[3^{4(x-y)}=3^1\] Taking the logarithm base 3, we get 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 by
Problem 17 		Followed by
Problem 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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