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 "2003 AMC 12A Problems/Problem 24"
 
(9 intermediate revisions by 5 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
 
 
If <math>a\geq b > 1,</math> what is the largest possible value of <math>\log_{a}(a/b) + \log_{b}(b/a)?</math>
 
If <math>a\geq b > 1,</math> what is the largest possible value of <math>\log_{a}(a/b) + \log_{b}(b/a)?</math>
  
Line 12: Line 11:
  
 
== Solution ==
 
== Solution ==
−
 
 
Using logarithmic rules, we see that
 
Using logarithmic rules, we see that
  
 
<cmath>\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)</cmath>
 
<cmath>\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)</cmath>
−
<cmath>=2-(\log_{a}b+\frac {1}{\log_{a}b})</cmath>
+
<cmath>=2-\left(\log_{a}b+\frac {1}{\log_{a}b}\right)</cmath>
  
−
Since <math>a</math> and <math>b</math> are both positive, using [[AM-GM]] gives that the term in parentheses must be at least <math>2</math>, so the largest possible values is <math>2-2=0 \Rightarrow \boxed{\textbf{B}}.</math>
+
Since <math>a</math> and <math>b</math> are both greater than <math>1</math>, using [[AM-GM]] gives that the term in parentheses must be at least <math>2</math>, so the largest possible values is <math>2-2=0 \Rightarrow \boxed{\textbf{B}}.</math>
  
 
Note that the maximum occurs when <math>a=b</math>.
 
Note that the maximum occurs when <math>a=b</math>.
 +
 +
 +
==Solution 2 (More Algebraic)==
 +
Similar to the previous solution, we use our logarithmic rules and start off the same way.
 +
<cmath>\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)</cmath>
 +
 +
However, now, we use a different logarithmic rule stating that <math>\log_{a}b</math> is simply equal to <math>\frac {\log_{10}b}{\log_{10}a}</math>. With this, we can rewrite our previous equation to give us <cmath>2-(\log_{a}b+\log_{b}a) = 2 - \left(\frac {\log_{10}b}{\log_{10}a} + \frac {\log_{10}a}{\log_{10}b}\right)</cmath>.
 +
 +
We can now cross multiply to get that <cmath>2 - \left(\frac {\log_{10}b}{\log_{10}a} + \frac {\log_{10}a}{\log_{10}b}\right) = 2 - \left(\frac {2 \cdot \log_{10}b \cdot \log_{10}a}{\log_{10}b \cdot \log_{10}a}\right)</cmath> Finally, we cancel to get <math>2-2=0 \Rightarrow \boxed{\textbf{(B) 0}}.</math>
 +
 +
Solution by: armang32324
 +
 +
==Solution 3 (Calculus)==
 +
By the logarithmic rules, we have <math>2-(\log_a{b}+\frac{1}{\log_a{b}})</math>.
 +
Let <math>x=\log_a{b}</math>. Thus, the expression becomes <math>2-(x+\frac{1}{x})</math>. We want to find the maximum of the function. To do so, we will find its derivative and let it equal to 0. We get: <math>\frac{d}{dx}\big(2-(x+\frac{1}{x})\big)=\frac{d}{dx}\big(2-x-\frac{1}{x})=-1+x^{-2}=0 \implies \frac{1}{x^2}=1, x^2=1, x=\pm1.</math>
 +
Since <math>a\geq b>1, x=-1</math> is not a solution. Thus, <math>x=1</math>. Substituting it into the original expression <math>2-(x+\frac{1}{x})</math>, we get <math>2-(1+\frac{1}{1})=2-2=\boxed{0}</math>.
  
 
==Video Solution==
 
==Video Solution==
  
 
The Link: https://www.youtube.com/watch?v=InF2phZZi2A&t=1s
 
The Link: https://www.youtube.com/watch?v=InF2phZZi2A&t=1s
 +
 +
-MistyMathMusic
  
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2003|ab=A|num-b=23|num-a=25}}
 
{{AMC12 box|year=2003|ab=A|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 08:35, 6 September 2023

Problem

If $a\geq b > 1,$ what is the largest possible value of $\log_{a}(a/b) + \log_{b}(b/a)?$

$\mathrm{(A)}\ -2 \qquad \mathrm{(B)}\ 0 \qquad \mathrm{(C)}\ 2 \qquad \mathrm{(D)}\ 3 \qquad \mathrm{(E)}\ 4$

Solution

Using logarithmic rules, we see that

\[\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)\] \[=2-\left(\log_{a}b+\frac {1}{\log_{a}b}\right)\]

Since $a$ and $b$ are both greater than $1$, using AM-GM gives that the term in parentheses must be at least $2$, so the largest possible values is $2-2=0 \Rightarrow \boxed{\textbf{B}}.$

Note that the maximum occurs when $a=b$.


Solution 2 (More Algebraic)

Similar to the previous solution, we use our logarithmic rules and start off the same way. \[\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)\]

However, now, we use a different logarithmic rule stating that $\log_{a}b$ is simply equal to $\frac {\log_{10}b}{\log_{10}a}$. With this, we can rewrite our previous equation to give us \[2-(\log_{a}b+\log_{b}a) = 2 - \left(\frac {\log_{10}b}{\log_{10}a} + \frac {\log_{10}a}{\log_{10}b}\right)\].

We can now cross multiply to get that \[2 - \left(\frac {\log_{10}b}{\log_{10}a} + \frac {\log_{10}a}{\log_{10}b}\right) = 2 - \left(\frac {2 \cdot \log_{10}b \cdot \log_{10}a}{\log_{10}b \cdot \log_{10}a}\right)\] Finally, we cancel to get $2-2=0 \Rightarrow \boxed{\textbf{(B) 0}}.$

Solution by: armang32324

Solution 3 (Calculus)

By the logarithmic rules, we have $2-(\log_a{b}+\frac{1}{\log_a{b}})$. Let $x=\log_a{b}$. Thus, the expression becomes $2-(x+\frac{1}{x})$. We want to find the maximum of the function. To do so, we will find its derivative and let it equal to 0. We get: $\frac{d}{dx}\big(2-(x+\frac{1}{x})\big)=\frac{d}{dx}\big(2-x-\frac{1}{x})=-1+x^{-2}=0 \implies \frac{1}{x^2}=1, x^2=1, x=\pm1.$ Since $a\geq b>1, x=-1$ is not a solution. Thus, $x=1$. Substituting it into the original expression $2-(x+\frac{1}{x})$, we get $2-(1+\frac{1}{1})=2-2=\boxed{0}$.

Video Solution

The Link: https://www.youtube.com/watch?v=InF2phZZi2A&t=1s

-MistyMathMusic

See Also

2003 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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
All AMC 12 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=2003_AMC_12A_Problems/Problem_24&oldid=197907"