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

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== Solution ==
 
== Solution ==
<math>\fbox{B}</math>
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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>.
  
 
== See also ==
 
== See also ==

Latest revision as of 18:58, 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 $\boxed{B}$.

See also

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