Difference between revisions of "1988 AHSME Problems/Problem 26"
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==Solution== | ==Solution== | ||
We can rewrite the equation as <math>\frac{\log{p}}{\log{9}} = \frac{\log{q}}{\log{12}} = \frac{\log{(p + q)}}{\log{16}}</math>. Then, the system can be split into 3 pairs: <math>\frac{\log{p}}{\log{9}} = \frac{\log{q}}{\log{12}}</math>, <math>\frac{\log{q}}{\log{12}} = \frac{\log{(p + q)}}{\log{16}}</math>, and <math>\frac{\log{p}}{\log{9}} = \frac{\log{(p + q)}}{\log{16}}</math>. Cross-multiplying in the first two, we obtain: <cmath>(\log{12})(\log{p}) = (2\log{3})(\log{q})</cmath> and <cmath>(\log{12})(\log{(p + q)}) = (2\log{4})(\log{q})</cmath> | We can rewrite the equation as <math>\frac{\log{p}}{\log{9}} = \frac{\log{q}}{\log{12}} = \frac{\log{(p + q)}}{\log{16}}</math>. Then, the system can be split into 3 pairs: <math>\frac{\log{p}}{\log{9}} = \frac{\log{q}}{\log{12}}</math>, <math>\frac{\log{q}}{\log{12}} = \frac{\log{(p + q)}}{\log{16}}</math>, and <math>\frac{\log{p}}{\log{9}} = \frac{\log{(p + q)}}{\log{16}}</math>. Cross-multiplying in the first two, we obtain: <cmath>(\log{12})(\log{p}) = (2\log{3})(\log{q})</cmath> and <cmath>(\log{12})(\log{(p + q)}) = (2\log{4})(\log{q})</cmath> | ||
| − | Adding these equations results in: <cmath>(\log{12})(\log{p(p+q)}) = (2\log{12})(\log{q})</cmath> which simplifies to <cmath>p(p + q) = q^2</cmath> Dividing by <math>pq</math> on both sides gives: <math>\frac{p+q}{q} = \frac{q}{p} = \frac{p}{q} + 1</math>. We set the desired value, <math>q/p</math> to <math>x</math> and substitute it into our equation: <math>\frac{1}{x} + 1 = x \implies x^2 - x - 1 = 0</math> which is solved to get our answer: <math>\boxed{\text{(C) } \frac{1 + \sqrt{5}}{2}}</math>. | + | Adding these equations results in: <cmath>(\log{12})(\log{p(p+q)}) = (2\log{12})(\log{q})</cmath> which simplifies to <cmath>p(p + q) = q^2</cmath> Dividing by <math>pq</math> on both sides gives: <math>\frac{p+q}{q} = \frac{q}{p} = \frac{p}{q} + 1</math>. We set the desired value, <math>q/p</math> to <math>x</math> and substitute it into our equation: <math>\frac{1}{x} + 1 = x \implies x^2 - x - 1 = 0</math> which is solved to get our answer: <math>\boxed{\text{(C) } \frac{1 + \sqrt{5}}{2}}</math>. -lucasxia01 |
== See also == | == See also == | ||
Revision as of 23:28, 7 June 2016
Problem
Suppose that 
and 
are positive numbers for which
What is the value of 
?
Solution
We can rewrite the equation as 
. Then, the system can be split into 3 pairs: 
, 
, and 
. Cross-multiplying in the first two, we obtain: ![\[(\log{12})(\log{p}) = (2\log{3})(\log{q})\]](http://latex.artofproblemsolving.com/7/0/d/70d68fdccdde58bd44ca734fc8bb563e4204b058.png)
and ![\[(\log{12})(\log{(p + q)}) = (2\log{4})(\log{q})\]](http://latex.artofproblemsolving.com/8/f/a/8fa4de9df972754f9290d1a908e61ef01c942d54.png)
Adding these equations results in: ![\[(\log{12})(\log{p(p+q)}) = (2\log{12})(\log{q})\]](http://latex.artofproblemsolving.com/b/e/f/befe925e52672f5c6321f6dbb74534f3b94e7f68.png)
which simplifies to ![\[p(p + q) = q^2\]](http://latex.artofproblemsolving.com/5/8/5/585e217228a272b5f344d72599861f9dc9cebd2f.png)
Dividing by 
on both sides gives: 
. We set the desired value, 
to 
and substitute it into our equation: 
which is solved to get our answer: 
. -lucasxia01
See also
| 1988 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 25 |
Followed by Problem 27 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
