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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== See also == | == See also == |
Latest revision as of 17:58, 4 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 .
See also
1966 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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All AHSME Problems and Solutions |
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