Difference between revisions of "1953 AHSME Problems/Problem 39"
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| + | ==Problem== | ||
| + | |||
| + | The product, <math>\log_a b \cdot \log_b a</math> is equal to: | ||
| + | |||
| + | <math>\textbf{(A)}\ 1 \qquad | ||
| + | \textbf{(B)}\ a \qquad | ||
| + | \textbf{(C)}\ b \qquad | ||
| + | \textbf{(D)}\ ab \qquad | ||
| + | \textbf{(E)}\ \text{none of these} </math> | ||
| + | |||
| + | ==Solution== | ||
<cmath>a^x=b</cmath> | <cmath>a^x=b</cmath> | ||
<cmath>b^y=a</cmath> | <cmath>b^y=a</cmath> | ||
| Line 4: | Line 15: | ||
<cmath>xy=1</cmath> | <cmath>xy=1</cmath> | ||
<cmath>\log_a b\log_b a=1</cmath> | <cmath>\log_a b\log_b a=1</cmath> | ||
| − | As a result, the answer should be 1 \boxed{A}. | + | As a result, the answer should be <math>\boxed{\textbf{(A) }1}</math>. |
| + | |||
| + | ==Solution 2== | ||
| + | Apply the change of base formula to <math>\log_a b</math> and <math>\log_b a</math>. For simplicity, let us convert to base-10 log. | ||
| + | By change of base, the expression becomes <math>\frac{\log b}{\log a} * \frac{\log a}{\log b} = \boxed{\textbf{(A) }1}</math>. | ||
==See Also== | ==See Also== | ||
Latest revision as of 16:57, 22 April 2020
Contents
Problem
The product, 
is equal to:
Solution
![\[a^x=b\]](http://latex.artofproblemsolving.com/f/3/7/f375b04de3c53239b27b6d7abad77ded50419ae2.png)
![\[b^y=a\]](http://latex.artofproblemsolving.com/9/f/1/9f199df2a6672f2c121f76bbbbf16d857f7bcb3d.png)
![\[{a^x}^y=a\]](http://latex.artofproblemsolving.com/8/3/a/83a9dc65c79b012c6b82784abe87431cf6adf1c8.png)
![\[xy=1\]](http://latex.artofproblemsolving.com/4/9/7/4979305dd5b73b3790d6c4b89d10826fe6a6bbbc.png)
![\[\log_a b\log_b a=1\]](http://latex.artofproblemsolving.com/0/3/0/0307e46f37d6ad324959095283c8c5aa5f4de44a.png)
As a result, the answer should be 
.
Solution 2
Apply the change of base formula to 
and 
. For simplicity, let us convert to base-10 log.
By change of base, the expression becomes 
.
See Also
| 1953 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 38 |
Followed by Problem 40 | |
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| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
