Difference between revisions of "1953 AHSME Problems/Problem 39"

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==Problem==
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The product, <math>\log_a b \cdot \log_b a</math> is equal to:
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<math>\textbf{(A)}\ 1 \qquad
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\textbf{(B)}\ a \qquad
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\textbf{(C)}\ b \qquad
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\textbf{(D)}\ ab \qquad
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\textbf{(E)}\ \text{none of these}    </math>
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==Solution==
 
<cmath>a^x=b</cmath>
 
<cmath>a^x=b</cmath>
 
<cmath>b^y=a</cmath>
 
<cmath>b^y=a</cmath>

Revision as of 00:44, 4 February 2020

Problem

The product, $\log_a b \cdot \log_b a$ is equal to:

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ a \qquad \textbf{(C)}\ b \qquad \textbf{(D)}\ ab \qquad \textbf{(E)}\ \text{none of these}$

Solution

\[a^x=b\] \[b^y=a\] \[{a^x}^y=a\] \[xy=1\] \[\log_a b\log_b a=1\] As a result, the answer should be 1 \boxed{A}.

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 38
Followed by
Problem 40
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 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions


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