Difference between revisions of "1966 AHSME Problems/Problem 24"

Revision as of 15:48, 2 January 2020

Problem

If $Log_M{N}=Log_N{M},M \ne N,MN>0,M \ne 1, N \ne 1$ , then $MN$ equals:

$\text{(A) } \frac{1}{2} \quad \text{(B) } 1 \quad \text{(C) } 2 \quad \text{(D) } 10 \\ \text{(E) a number greater than 2 and less than 10}$

Solution

If we change the base of $Log_M{N}$ to base $N$ , we get $\frac{1}{log_N{M}}= Log_N{M}$ . Multiplying both sides by $Log_N{M}$ , we get $Log_N{M}^2=1$ . Since $N\not = M$ , $Log_N{M}=-1$ . So $N^{-1}=M \rightarrow \frac{1}{N} = M \rightarrow MN=1$ . So the answer is $\boxed{B}$ .

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.

@@ Line 5: / Line 5: @@
 == Solution ==
-<math>\fbox{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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Difference between revisions of "1966 AHSME Problems/Problem 24"

Revision as of 15:48, 2 January 2020

Problem

Solution

See also