1953 AMC 12 Problem 39

Attempt Problem 39 of the 1953 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1953 AMC 12 solutions, or check the answer key.

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39.

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

11

aa

bb

abab

none of these

Answer: A
Concepts:logarithmschange of base
Difficulty rating: 1180
Small Hint:

Use the change-of-base formula on both logarithms

Big Hint:

The two resulting fractions are reciprocals

Solution:

For permissible bases and arguments, logablogba=logblogalogalogb=1. \begin{aligned} \log_a b\cdot\log_b a &=\frac{\log b}{\log a}\\ &\quad{}\cdot\frac{\log a}{\log b}\\ &=1. \end{aligned}

Thus, the correct answer is A.

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Problem 39 in Other Years

1950 AMC 12 · 1951 AMC 12 · 1952 AMC 12 · 1954 AMC 12 · 1955 AMC 12 · 1956 AMC 12 · 1957 AMC 12 · 1958 AMC 12 · 1959 AMC 12 · 1960 AMC 12 · 1961 AMC 12 · 1962 AMC 12 · 1963 AMC 12 · 1964 AMC 12 · 1965 AMC 12 · 1966 AMC 12 · 1967 AMC 12