Problem 42 - Calculus and its applications

Question

Given \(\ln 4=1.3863\) and \(\ln 5=1.6094,\) find each value. Do not use a calculator. $$ \ln \left(\frac{e}{5}\right) $$

Step-by-Step Solution

Answer

\( \ln \left(\frac{e}{5}\right) = -0.6094 \).

Step 1: Apply Property of Logarithms

Use the property \( \ln(a/b) = \ln a - \ln b \). In this case, we have \( a = e \) and \( b = 5 \). Thus, \( \ln \left(\frac{e}{5}\right) = \ln e - \ln 5 \).

Step 2: Substitute Known Values

From the properties of logarithms, we know \( \ln e = 1 \). Substitute this and the given value \( \ln 5 = 1.6094 \) into the equation to find \( \ln \left(\frac{e}{5}\right) = 1 - 1.6094 \).

Step 3: Perform the Calculation

Subtract the logarithms: \( \ln \left(\frac{e}{5}\right) = 1 - 1.6094 = -0.6094 \).

Key Concepts

Properties of Logarithms

Logarithms have several unique properties that allow us to simplify complex logarithmic expressions. These properties are rules or laws related to how logarithms work. Understanding and applying these properties can make solving logarithmic equations much simpler. Here are some key properties:

• Product Property: This property states that the log of a product is the sum of the logs. Mathematically, it’s expressed as \( \log_b(mn) = \log_b(m) + \log_b(n) \).
• Quotient Property: This property is useful when you have a division within the logarithm. It states that the log of a quotient is the difference of the logs. It is given by \( \log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n) \).
• Power Property: This property shows how to deal with exponents inside a log. The log of a number raised to a power can be brought down like this: \( \log_b(m^n) = n \log_b(m) \).

In the original exercise, we used the Quotient Property to turn the expression \( \ln\left(\frac{e}{5}\right) \) into \( \ln e - \ln 5 \). This simplification is a fundamental step in solving logarithmic problems.

Natural Logarithm

A natural logarithm is a logarithm to the base \( e \), where \( e \approx 2.71828 \). It is denoted by \( \ln \), which stands for 'logarithm natural.' Natural logarithms are incredibly prevalent in various fields such as mathematics, physics, and engineering because of the unique properties of the number \( e \).

The natural logarithm is particularly important because:

• It relates closely to exponential growth and decay, which appear frequently in natural processes.
• The derivative of \( e^x \) is \( e^x \), which simplifies many calculations in calculus.
• Whenever you see \( \ln e \), like in our exercise, it equals 1. That’s a fundamental fact that often simplifies many logarithmic calculations.

For our problem, knowing that \( \ln e = 1 \) was crucial. It allowed us to substitute \( 1 \) for \( \ln e \) in our calculation, making the process straightforward.

Logarithmic Calculations

Performing logarithmic calculations may seem daunting at first, but once you understand the properties and know the basic values, like \( \ln e \), it becomes much more accessible. When given values and equations, the key lies in recognizing which logarithmic properties to apply so the problem simplifies naturally.

In our problem, calculating \( \ln \left(\frac{e}{5}\right) \) without a calculator involved a few important steps:

• Recognize the Form: Image the problem in terms of known logarithmic properties, such as the quotient rule used here.
• Substitute Known Values: Use known values and basic logarithmic understandings like \( \ln e = 1 \) to reduce the expression.
• Perform Algebraic Operations: After substitution, execute the simple arithmetic operations left over, such as the subtraction seen in our example, \( 1 - 1.6094 \).

Through practice and familiarity with these steps, logarithmic calculations can become straightforward, as demonstrated by our ability to find \( \ln \left(\frac{e}{5}\right) \) without needing a calculator.