Difference between revisions of "1953 AHSME Problems/Problem 39"
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<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 <math>\boxed{\textbf{(A) }1}</math>. | As a result, the answer should be <math>\boxed{\textbf{(A) }1}</math>. | ||
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+ | ==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
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 |
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