Problem 1 - Thomas Calculus

Question

Express the following logarithms in terms of \(\ln 2\) and \(\ln 3 .\) $$ \begin{array}{lll}{\text { a. } \ln 0.75} & {\text { b. } \ln (4 / 9)} & {\text { c. } \ln (1 / 2)} \\ {\text { d. } \ln \sqrt[3]{9}} & {\text { e. } \ln 3 \sqrt{2}} & {\text { f. } \ln \sqrt{13.5}}\end{array} $$

Step-by-Step Solution

Answer

a. \(\ln 3 - 2\ln 2\); b. \(2\ln 2 - 2\ln 3\); c. \(-\ln 2\); d. \(\frac{2}{3} \ln 3\); e. \(\ln 3 + \frac{1}{2} \ln 2\); f. \(\frac{3}{2} \ln 3 - \frac{1}{2} \ln 2\).

Step 1: Simplify \(\ln 0.75\)

We start by expressing 0.75 as a fraction: \( 0.75 = \frac{3}{4} \). Then, use the property \( \ln \frac{a}{b} = \ln a - \ln b \). This gives us \( \ln 0.75 = \ln 3 - \ln 4 \). Since \( 4 = 2^2 \), we have \( \ln 4 = 2 \ln 2 \). Therefore, \( \ln 0.75 = \ln 3 - 2 \ln 2 \).

Step 2: Simplify \(\ln (4/9)\)

Express \( \frac{4}{9} \) as a fraction. Use \( \ln \frac{a}{b} = \ln a - \ln b \). So, \( \ln \frac{4}{9} = \ln 4 - \ln 9 \). Convert \( \ln 4 \) to \( 2\ln 2 \) and \( \ln 9 \) to \( 2\ln 3 \). Thus, \( \ln \frac{4}{9} = 2\ln 2 - 2\ln 3 \).

Step 3: Simplify \(\ln (1/2)\)

Use the property \( \ln \frac{1}{a} = -\ln a \). So, \( \ln \frac{1}{2} = -\ln 2 \). This means the expression is simply \( -\ln 2 \).

Step 4: Simplify \(\ln \sqrt[3]{9}\)

Express \( \sqrt[3]{9} \) using exponents: \( 9^{1/3} \). Then use the property \( \ln a^b = b \ln a \). This gives \( \ln 9^{1/3} = \frac{1}{3} \ln 9 \). Since \( 9 = 3^2 \), \( \ln 9 = 2 \ln 3 \), so \( \ln 9^{1/3} = \frac{2}{3} \ln 3 \).

Step 5: Simplify \(\ln 3 \sqrt{2}\)

Express \( 3 \sqrt{2} \) as \( 3 \times 2^{1/2} \). Use \( \ln(ab) = \ln a + \ln b \). Therefore, \( \ln(3\sqrt{2}) = \ln 3 + \ln 2^{1/2} = \ln 3 + \frac{1}{2} \ln 2 \).

Step 6: Simplify \(\ln \sqrt{13.5}\)

Express \( \sqrt{13.5} \) as \( (13.5)^{1/2} \). Use \( \ln a^b = b \ln a \). Therefore, \( \ln (13.5)^{1/2} = \frac{1}{2} \ln 13.5 \). Rewrite 13.5 as \( \frac{27}{2} \), so \( \ln 13.5 = \ln \frac{27}{2} = \ln 27 - \ln 2 \). Since \( \ln 27 = 3 \ln 3 \), we have \( \ln 13.5 = 3\ln 3 - \ln 2 \). Finally, \( \ln \sqrt{13.5} = \frac{1}{2}(3\ln 3 - \ln 2) = \frac{3}{2} \ln 3 - \frac{1}{2} \ln 2 \).

Key Concepts

Logarithm Properties

Logarithms have a set of properties that make them very versatile tools in mathematics. By understanding and applying these properties, complex logarithmic expressions can be simplified systematically. Here are some essential properties that are frequently used:

• Product Property: \( \ln(ab) = \ln a + \ln b \). This property states that the logarithm of a product is the sum of the logarithms.
• Quotient Property: \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \). The logarithm of a quotient is the difference of the logarithms.
• Power Property: \( \ln(a^b) = b \ln a \). This property states that the logarithm of a power is the exponent times the logarithm.

Understanding these properties is fundamental to simplifying logarithmic expressions. Each property helps to break down complex terms into simpler terms that are more manageable.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a logarithm with the base \( e \) (approximately 2.718). This form of logarithm frequently appears in calculus, real-world exponential growth processes, and complex calculations.
Here are a few characteristics of the natural logarithm:

• \( \ln(e) = 1 \) because \( e^1 = e \).
• \( \ln(1) = 0 \) because any number to the power of zero equals one.

The natural logarithm simplifies managing calculations involving exponentials, such as continuous compound interest and radioactive decay. It is an essential tool in advanced mathematics, particularly in calculus.

Simplifying Expressions

Simplifying logarithmic expressions often involves using logarithm properties to write complex expressions in terms of simpler ones. This can make it easier to understand and solve mathematical and real-world problems.
To simplify, you should:

• Identify forms that match one of the logarithmic properties (product, quotient, power).
• Use these properties to simplify the logarithm into basic terms that can further be calculated or interpreted.
• Substitute known values or relationships to express the logarithm in a more straightforward form.

Successfully simplifying expressions allows one to convert logarithmic terms into base elements like \( \ln 2 \) and \( \ln 3 \), as shown in the original problem.

Change of Base Formula

The change of base formula is a useful tool for converting logarithms from one base to another, though in this exercise it's not directly used to express in terms of \( \ln 2 \) and \( \ln 3 \). However, it helps in understanding the flexibility of logarithms.
The formula is:

• For a logarithm \( \log_b a \), the change of base can be expressed as \( \frac{\ln a}{\ln b} \).
• This formula allows converting any logarithm to a natural logarithm (or to any other base), making calculating them on a standard calculator easier.

Understanding this formula broadens your understanding of how logarithms can be manipulated and used in different mathematical scenarios.