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

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(Solution)
 
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<cmath>\log_a b\log_b a=1</cmath>
 
<cmath>\log_a b\log_b a=1</cmath>
 
As a result, the answer should be <math>\boxed{\textbf{(A) }1}</math>.
 
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 17:57, 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 38
Followed by
Problem 40
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All AHSME Problems and Solutions


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