Difference between revisions of "1966 AHSME Problems/Problem 24"
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== Solution == | == Solution == | ||
If we change the base of <math>Log_M{N}</math> to base <math>N</math>, we get <math>\frac{1}{log_N{M}}= Log_N{M}</math>. Multiplying both sides by <math>Log_N{M}</math>, we get <math>Log_N{M}^2=1</math>. Since <math>N\not = M</math>, <math>Log_N{M}=-1</math>. So <math>N^{-1}=M \rightarrow \frac{1}{N} = M \rightarrow MN=1</math>. So the answer is <math>\boxed{B}</math>. | If we change the base of <math>Log_M{N}</math> to base <math>N</math>, we get <math>\frac{1}{log_N{M}}= Log_N{M}</math>. Multiplying both sides by <math>Log_N{M}</math>, we get <math>Log_N{M}^2=1</math>. Since <math>N\not = M</math>, <math>Log_N{M}=-1</math>. So <math>N^{-1}=M \rightarrow \frac{1}{N} = M \rightarrow MN=1</math>. So the answer is <math>\boxed{B}</math>. | ||
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| + | ~Ultraman | ||
== See also == | == See also == | ||
Revision as of 15:49, 2 January 2020
Problem
If 
, then 
equals:
Solution
If we change the base of 
to base 
, we get 
. Multiplying both sides by 
, we get 
. Since 
, 
. So 
. So the answer is 
.
~Ultraman
See also
| 1966 AHSME (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 • 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. 
