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

Revision as of 19:23, 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 $\boxed{\textbf{(E)}}$ .

1960 AHSME (Problems • Answer Key • Resources)
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All AHSME Problems and Solutions

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−	== Problem 18==	+	==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:

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

Revision as of 19:23, 9 May 2018

Problem

Solution

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