Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1988 AHSME Problems/Problem 26"

m (→‎Solution)
(→‎Solution 2)
 
(6 intermediate revisions by 4 users not shown)
Line 1: Line 1:
 
==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>?
−
 
 
−
<math>\log_{9}(p) = \log_{12}(q) = \log_{16}(p+q)</math>
 
−
 
 
−
What is the value of <math>\frac{q}{p}</math>?
 
  
 
<math>\textbf{(A)}\ \frac{4}{3}\qquad
 
<math>\textbf{(A)}\ \frac{4}{3}\qquad
Line 14: Line 10:
  
  
−
==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{(D) } \frac{1 + \sqrt{5}}{2}}</math>. -lucasxia01
 
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
 +
 +
==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 11:15, 6 August 2023

Problem

Suppose that $p$ and $q$ are positive numbers for which \[\operatorname{log}_{9}(p) = \operatorname{log}_{12}(q) = \operatorname{log}_{16}(p+q).\] What is the value of $\frac{q}{p}$?

$\textbf{(A)}\ \frac{4}{3}\qquad \textbf{(B)}\ \frac{1+\sqrt{3}}{2}\qquad \textbf{(C)}\ \frac{8}{5}\qquad \textbf{(D)}\ \frac{1+\sqrt{5}}{2}\qquad \textbf{(E)}\ \frac{16}{9}$


Solution 1

We can rewrite the equation as $\frac{\log{p}}{\log{9}} = \frac{\log{q}}{\log{12}} = \frac{\log{(p + q)}}{\log{16}}$. Then, the system can be split into 3 pairs: $\frac{\log{p}}{\log{9}} = \frac{\log{q}}{\log{12}}$, $\frac{\log{q}}{\log{12}} = \frac{\log{(p + q)}}{\log{16}}$, and $\frac{\log{p}}{\log{9}} = \frac{\log{(p + q)}}{\log{16}}$. Cross-multiplying in the first two, we obtain: \[(\log{12})(\log{p}) = (2\log{3})(\log{q})\] and \[(\log{12})(\log{(p + q)}) = (2\log{4})(\log{q})\] Adding these equations results in: \[(\log{12})(\log{p(p+q)}) = (2\log{12})(\log{q})\] which simplifies to \[p(p + q) = q^2\] Dividing by $pq$ on both sides gives: $\frac{p+q}{q} = \frac{q}{p} = \frac{p}{q} + 1$. We set the desired value, $q/p$ to $x$ and substitute it into our equation: $\frac{1}{x} + 1 = x \implies x^2 - x - 1 = 0$ which is solved to get our answer: $\boxed{\text{(D) } \frac{1 + \sqrt{5}}{2}}$. -lucasxia01

Solution 2

For some number t:

$p = 9^{t}$

$q = 12^{t}$

$p + q = 16^{t}$

Next we can divide $p + q$ by $p$ to obtain $\frac{p+q}{p} = 1 + \frac{q}{p}$

Furthermore, we know that

$\frac{p+q}{p} = (\frac{16}{9})^{t}$ and $\frac{q}{p} = (\frac{4}{3})^{t}$

Substituting into the previous equation, we get $(\frac{16}{9})^{t} = 1 + (\frac{4}{3})^{t}$

Let $x = (\frac{4}{3})^{t}$ and we can observe that $x = \frac{q}{p}$, then similarly to solution 1: $x^2 = 1 + x$, in which we get: $\boxed{\text{(D) } \frac{1 + \sqrt{5}}{2}}$ - 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1988_AHSME_Problems/Problem_26&oldid=196487"