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

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<cmath>xy=1</cmath>
 
<cmath>xy=1</cmath>
 
<cmath>\log_a b\log_b a=1</cmath>
 
<cmath>\log_a b\log_b a=1</cmath>
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As a result, the answer should be 1 \boxed{A}.
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As a result, the answer should be <math>\boxed{\textbf{(A) }1}</math>.
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==Solution 2==
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Apply the change of base formula to  <math>\log_a b</math> and <math>\log_b a</math>. For simplicity, let us convert to base-10 log.
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By change of base, the expression becomes <math>\frac{\log b}{\log a} * \frac{\log a}{\log b} = \boxed{\textbf{(A) }1}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 16:57, 22 April 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 $\boxed{\textbf{(A) }1}$.

Solution 2

Apply the change of base formula to $\log_a b$ and $\log_b a$. For simplicity, let us convert to base-10 log. By change of base, the expression becomes $\frac{\log b}{\log a} * \frac{\log a}{\log b} = \boxed{\textbf{(A) }1}$.

See Also

1953 AHSC (Problems • Answer Key • Resources)
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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