Ch. 1 - Functions

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 1 - Functions

Problem 1.20

Chapter 1, Problem 1.20

For a certain constant a>1, ln a≈3.8067 . Find approximate values of log₂ a and logₐ 2 using the fact that ln 2≈0.6931.

Verified step by step guidance

Use the change of base formula for logarithms: \( \log_b a = \frac{\ln a}{\ln b} \).

To find \( \log_2 a \), apply the formula: \( \log_2 a = \frac{\ln a}{\ln 2} \).

Substitute the given values: \( \ln a \approx 3.8067 \) and \( \ln 2 \approx 0.6931 \).

Calculate \( \log_2 a \) using the substituted values: \( \log_2 a = \frac{3.8067}{0.6931} \).

To find \( \log_a 2 \), use the reciprocal property of logarithms: \( \log_a 2 = \frac{1}{\log_2 a} \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Change of Base Formula

The change of base formula allows us to convert logarithms from one base to another. It states that log_b(x) = log_k(x) / log_k(b) for any positive k. This is particularly useful when we need to compute logarithms in a base that is not readily available, such as converting natural logarithms (ln) to base 2 or base a.

Natural Logarithm (ln)

The natural logarithm, denoted as ln, is the logarithm to the base e, where e is approximately 2.71828. It is a fundamental concept in calculus and is used extensively in various applications, including growth models and compound interest. In this problem, ln a and ln 2 are provided, which are essential for calculating log₂ a and logₐ 2.

Approximation of Logarithmic Values

In this context, approximating logarithmic values involves using known logarithmic values to estimate others. Given ln a and ln 2, we can derive log₂ a and logₐ 2 using the change of base formula. This method is particularly useful when exact values are difficult to compute, allowing for practical estimations in mathematical problems.

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