Difference between revisions of "1988 AHSME Problems/Problem 26"
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==Problem== | ==Problem== | ||
− | Suppose that <math>p</math> and <math>q</math> are positive numbers for which | + | Suppose that <math>p</math> and <math>q</math> are positive numbers for which <cmath>\operatorname{log}_{9}(p) = \operatorname{log}_{12}(q) = \operatorname{log}_{16}(p+q).</cmath> What is the value of <math>\frac{q}{p}</math>? |
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− | < | ||
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− | What is the value of <math>\frac{q}{p}</math>? | ||
<math>\textbf{(A)}\ \frac{4}{3}\qquad | <math>\textbf{(A)}\ \frac{4}{3}\qquad | ||
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− | ==Solution== | + | ==Solution 1== |
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{( | + | 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{(D) } \frac{1 + \sqrt{5}}{2}}</math>. -lucasxia01 |
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+ | ==Solution 2== | ||
+ | For some number t: | ||
+ | |||
+ | <math>p = 9^{t}</math> | ||
+ | |||
+ | <math>q = 12^{t}</math> | ||
+ | |||
+ | <math>p + q = 16^{t}</math> | ||
+ | |||
+ | Next we can divide <math>p + q</math> by <math>p</math> to obtain | ||
+ | <math>\frac{p+q}{p} = 1 + \frac{q}{p}</math> | ||
+ | |||
+ | Furthermore, we know that | ||
+ | |||
+ | <math>\frac{p+q}{p} = (\frac{16}{9})^{t}</math> and <math>\frac{q}{p} = (\frac{4}{3})^{t}</math> | ||
+ | |||
+ | Substituting into the previous equation, we get <math>(\frac{16}{9})^{t} = 1 + (\frac{4}{3})^{t}</math> | ||
+ | |||
+ | Let <math>x = (\frac{4}{3})^{t}</math> and we can observe that <math> x = \frac{q}{p}</math>, then similarly to solution 1: <math>x^2 = 1 + x</math>, in which we get: <math>\boxed{\text{(D) } \frac{1 + \sqrt{5}}{2}}</math> - ehmmaq | ||
== See also == | == See also == |
Latest revision as of 12:15, 6 August 2023
Contents
Problem
Suppose that and
are positive numbers for which
What is the value of
?
Solution 1
We can rewrite the equation as . Then, the system can be split into 3 pairs:
,
, and
. Cross-multiplying in the first two, we obtain:
and
Adding these equations results in:
which simplifies to
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
Solution 2
For some number t:
Next we can divide by
to obtain
Furthermore, we know that
and
Substituting into the previous equation, we get
Let and we can observe that
, then similarly to solution 1:
, in which we get:
- ehmmaq
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 |
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