1953 AHSME Problems/Problem 39

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 $\frac{\log b}{\log a} * \frac{\log a}{\log b} = \boxed{\textbf{(A) }1}$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1953 AHSME Problems/Problem 39

Contents

Problem

Solution

Solution 2

See Also