Problem 48 - Elementary and Intermediate Algebra

Question

Evaluate each expression without using a calculator. $$ \ln \sqrt[5]{e} $$

Step-by-Step Solution

Answer

The expression evaluates to \( \frac{1}{5} \).

Step 1: Simplify the Expression

The expression given is \( \ln \sqrt[5]{e} \). Start by rewriting \( \sqrt[5]{e} \) as a power of \( e \). We know that \( \sqrt[5]{e} = e^{1/5} \). Therefore, the expression becomes \( \ln(e^{1/5}) \).

Step 2: Apply Logarithmic Identity

Now apply the logarithmic identity \( \ln(a^b) = b \ln a \). In this case, \( a = e \) and \( b = 1/5 \). So, \( \ln(e^{1/5}) = \frac{1}{5} \ln e \).

Step 3: Evaluate \( \ln e \)

Recall that \( \ln e = 1 \) because the natural logarithm of \( e \) is always 1. Substitute this into the expression from Step 2: \( \frac{1}{5} \ln e = \frac{1}{5} \times 1 = \frac{1}{5} \).

Key Concepts

Exponential Functions

Exponential functions are a fundamental concept in mathematics, where a variable is in the exponent. These functions are described by expressions like \(f(x) = a^x\), where \(a\) is a constant and \(x\) is the variable. In our context, the number \(e\), approximately equal to 2.718, is often used as the base of these functions.

Exponential functions are powerful because they model growth and decay processes found in various fields, such as biology, finance, and physics. They can represent anything from population growth to radioactive decay.

A special property of the exponential function is its continuous compound growth. For instance, \(e^x\) has the unique property that its derivative is itself, which means it grows at a rate proportional to its size.

• \(e^{a+b} = e^a \cdot e^b\)
• \((e^a)^b = e^{ab}\)
• \(e^0 = 1\)

Logarithmic Identities

Logarithms are the inverses of exponential functions. The natural logarithm, denoted \(\ln\), is the logarithm to the base \(e\). It helps us solve equations involving exponential functions by transforming them into linear equations.

One of the most used log identities is \(\ln(a^b) = b \cdot \ln a\). This identity allows us to take the exponent out as a multiplier. In our exercise, \(\ln(e^{1/5})\) became \(\frac{1}{5} \ln e\) due to this identity.

Remember some other key identities:

• \(\ln(1) = 0\) because \(e^0 = 1\)
• \(\ln(e) = 1\) because \(e^1 = e\)
• \(\ln(ab) = \ln a + \ln b\)
• \(\ln\left(\frac{a}{b}\right) = \ln a - \ln b\)

Understanding these identities will make it easier to work with logarithms in different expressions and solve complex logarithmic problems.

Simplifying Expressions

Simplifying expressions is an essential mathematical skill that helps in breaking down complex problems into manageable steps. By rewriting expressions, we can make them easier to evaluate or compare.

Consider our original expression \(\ln \sqrt[5]{e}\). The goal of simplification is to transform it into a cleaner and more straightforward form, such as \(\frac{1}{5}\) in this case.

The steps involved typically include:

• Rewriting roots and powers: \(\sqrt[5]{e}\) becomes \(e^{1/5}\).
• Using logarithmic identities to simplify further, such as taking exponents out with \(b \cdot \ln a\).
• Substituting known values: Since \(\ln e = 1\), it allows us to compute directly.

Mastering the simplification of mathematical expressions not only aids in calculation but also in enhancing our understanding of mathematical relationships.