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: Rewrite 0.75 in terms of fractions

0.75 can be expressed as \( \frac{3}{4} \). So, \( \ln(0.75) = \ln\left(\frac{3}{4}\right) \).

Step 2: Apply the logarithm quotient rule

Using the quotient rule for logarithms, \( \ln\left(\frac{3}{4}\right) = \ln(3) - \ln(4) \).

Step 3: Express \( \ln(4) \) in terms of \( \ln(2) \)

Since \( 4 = 2^2 \), \( \ln(4) = \ln(2^2) = 2\ln(2) \).

Step 4: Write \( \ln(0.75) \) in terms of \( \ln(2) \) and \( \ln(3) \)

Substitute the expression for \( \ln(4) \), so \( \ln(0.75) = \ln(3) - 2\ln(2) \).

Step 5: Solve part b, express 4/9 using fractions

\( \ln\left(\frac{4}{9}\right) = \ln(4) - \ln(9) \).

Step 6: Express \( \ln(9) \) in terms of \( \ln(3) \)

Since \( 9 = 3^2 \), \( \ln(9) = 2\ln(3) \).

Step 7: Express \( \ln(4/9) \) in terms of \( \ln(2) \) and \( \ln(3) \)

Substitute \( \ln(4) = 2\ln(2) \) and \( \ln(9) = 2\ln(3) \). Thus, \( \ln(4/9) = 2\ln(2) - 2\ln(3) \).

Step 8: Express 1/2 using fractions for Part c

\( \ln\left(\frac{1}{2}\right) = -\ln(2) \).

Step 9: Express \( \ln\sqrt[3]{9} \) for Part d

\( \ln\left(\sqrt[3]{9}\right) = \frac{1}{3}\ln(9) = \frac{1}{3}(2\ln(3)) = \frac{2}{3}\ln(3) \).

Step 10: Express \( \ln(\sqrt{3}^2) \) for Part e

\( \ln(3\sqrt{2}) = \ln(3) + \ln(\sqrt{2}) \).

Step 11: Simplify \( \ln(\sqrt{2}) \) for Part e

\( \ln(\sqrt{2}) = \frac{1}{2}\ln(2) \).

Step 12: Write \( \ln(3\sqrt{2}) \) in terms of \( \ln(2) \) and \( \ln(3) \)

Combine: \( \ln(3) + \frac{1}{2}\ln(2) \).

Step 13: Express \( \ln\sqrt{13.5} \) for Part f

\( \ln(\sqrt{13.5}) = \frac{1}{2}\ln(13.5) \).

Step 14: Rewrite 13.5 in terms of fractions

13.5 can be expressed as \( \frac{27}{2} \).

Step 15: Apply logarithm properties to \( \ln(\frac{27}{2}) \)

\( \ln(\frac{27}{2}) = \ln(27) - \ln(2) \).

Step 16: Express \( \ln(27) \) in terms of \( \ln(3) \)

Since \( 27 = 3^3 \), \( \ln(27) = 3\ln(3) \).

Step 17: Combine terms for \( \ln\sqrt{13.5} \)

\( \ln(\sqrt{13.5}) = \frac{1}{2}(3\ln(3) - \ln(2)) = \frac{3}{2}\ln(3) - \frac{1}{2}\ln(2) \).

Key Concepts

Logarithm Quotient Rule

The logarithm quotient rule is a fundamental property used to simplify logarithms of fractions. When you have the logarithm of a fraction, such as \( \ln\left( \frac{a}{b} \right) \), the rule states that it can be expressed as \( \ln(a) - \ln(b) \). This property helps to break down complex logarithmic expressions into simpler parts, making them easier to understand or calculate.
It essentially converts division inside the logarithm into subtraction between two logarithms. Using this rule is crucial when working with problems that involve fractions in the logarithmic form. In the original exercise, applying this rule allowed us to express \( \ln(0.75) \), \( \ln(4/9) \), and other similar fractions in terms of simpler logarithmic elements, helping us solve the exercise efficiently.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a special type of logarithm with the base \( e \), where \( e \approx 2.71828 \). It is widely used in mathematics due to its natural properties and simplicity in differentiation and integration.
Natural logarithms appear frequently in calculus and exponential growth problems. They are especially prevalent in exercises requiring the manipulation of powers and roots, like those found in the original problem. In the exercises, we often rewrite expressions in terms of natural logarithms involving known natural logs like \( \ln(2) \) and \( \ln(3) \), which simplifies much of the calculation.

• The natural logarithm is crucial for continuous growth models in real-world scenarios.
• It is utilized in solving differential equations involving exponential functions.

Logarithm Properties

Logarithm properties are rules that provide methods to alter and simplify logarithmic expressions. The fundamental properties include:

• The product rule: \( \ln(a \cdot b) = \ln(a) + \ln(b) \)
• The quotient rule: \( \ln\left( \frac{a}{b} \right) = \ln(a) - \ln(b) \)
• The power rule: \( \ln(a^b) = b \cdot \ln(a) \)

These properties are useful because they allow complex logarithmic expressions to be broken down into more manageable parts for calculation.
For instance, in the original exercise, utilizing properties like the power rule helped express numbers like \( 9 \) and \( 27 \) as powers of \( 3 \), simplifying expressions involving multiple terms. This makes it easier to express complex logarithmic terms equivalently with simpler expressions involving \( \ln(2) \) and \( \ln(3) \).

Simplifying Logarithms

Simplifying logarithms involves using logarithm properties to reduce an expression to its simplest form. The process includes:

• Rewriting complex numbers in terms of their factors or bases
• Applying logarithm properties like the product, quotient, and power rules
• Expressing results in terms of familiar logarithms, such as \( \ln(2) \) and \( \ln(3) \), as shown in the original exercise

This simplification is essential because it allows one to solve logarithms more easily by working with recognizable terms.
During simplification, understanding the factors of numbers becomes crucial. For instance, knowing that \( 4 = 2^2 \) and \( 27 = 3^3 \) allows the transformation of terms like \( \ln(4) \) and \( \ln(27) \) in terms of \( \ln(2) \) and \( \ln(3) \). This conversion simplifies the computations and provides a clearer pathway to the solution.