1966 AHSME Problems/Problem 24

Revision as of 15:48, 2 January 2020 by Ultraman (talk | contribs) (Solution)

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 $\boxed{B}$.

See also

1966 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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