Difference between revisions of "2003 AMC 12A Problems/Problem 24"

Revision as of 10:12, 30 June 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

Video Solution

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

-MistyMathMusic

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.

@@ Line 14: / Line 14: @@
 <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 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>
@@ Line 25: / Line 25: @@
 <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 - (\frac {\log_{10}b}{\log_{10}a} + \frac {\log_{10}a}{\log_{10}b})</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 - (\frac {\log_{10}b}{\log_{10}a} + \frac {\log_{10}a}{\log_{10}b}) = 2 - (\frac {2 * log_{10}b * \log_{10}a}{log_{10}b * \log_{10}a})</cmath> Finally, we cancel to get <math>2-2=0 \Rightarrow \boxed{\textbf{(B) 0}}.</math>
+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

Page

Toolbox

Search