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

Latest revision as of 17:58, 4 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
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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

Latest revision as of 17:58, 4 January 2020

Problem

Solution

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