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> | ||
| − | + | 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>. | ||
| + | ==See Also== | ||
| + | {{AHSME 40p box|year=1960|num-b=17|num-a=19}} | ||
| − | + | [[Category:Introductory Algebra Problems]] | |
| − | |||
Latest revision as of 21:14, 13 January 2023
Problem
The pair of equations 
and 
has:
Solution
Rewrite the equations so both sides have a common base.
![\[3^{x+y}=3^4\]](http://latex.artofproblemsolving.com/e/1/4/e14487119f4086e7ffa54c0f63278335430ede96.png)
![\[3^{4(x-y)}=3^1\]](http://latex.artofproblemsolving.com/9/f/5/9f5b09cdcfd2d07cbf76ed0e879c19e0453ca9b4.png)
Taking the logarithm base 3, we get a linear system of equations.
![\[x+y=4\]](http://latex.artofproblemsolving.com/9/a/7/9a7605802eb4d4200c9698e4ed6061549faaf01f.png)
![\[4x-4y=1\]](http://latex.artofproblemsolving.com/b/8/0/b8004b1cb80981f4c4126477e586827ae323602b.png)
Solve the system to get 
and 
. The answer is 
.
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 | ||
