Problem 46 - Essential Calculus Early Transcendentals

Question

Find the exact value of each expression (without a calculator.) $$ { (a) }e^{-2 \ln 5} \quad \text { (b) } \ln \left(\ln e^{2 n}\right) $$

Step-by-Step Solution

Answer

(a) \(\frac{1}{25}\); (b) \(\ln(2n)\).

Step 1: Simplify the Expression for Part (a)

We start with the expression \(e^{-2 \ln 5}\). By properties of logarithms and exponents, \(e^{\ln x} = x\). We can rewrite \(-2 \ln 5\) as \(\ln (5^{-2})\), leveraging the property \(a \ln b = \ln (b^a)\). Thus, the expression becomes \(e^{\ln (5^{-2})}\).

Step 2: Apply Exponential and Logarithmic Properties for Part (a)

Now, utilize the property \(e^{\ln x} = x\): applying this, \(e^{\ln (5^{-2})} = 5^{-2}\). Calculating \(5^{-2}\), we find \(\frac{1}{5^2} = \frac{1}{25}\). Thus, \(e^{-2 \ln 5} = \frac{1}{25}\).

Step 3: Simplify the Expression for Part (b)

We begin with \(\ln (\ln e^{2n})\). Start by simplifying the inner part, \(\ln e^{2n}\). The property \(\ln e^x = x\) simplifies this to \(2n\). Therefore, we need to evaluate \(\ln(2n)\).

Step 4: Final Evaluation for Part (b)

There is no further simplification available for \(\ln(2n)\) without specific value for \(n\). However, this is the simplified form using properties of logarithms. Therefore, the result for this expression remains \(\ln(2n)\).

Key Concepts

Logarithmic Properties

Logarithms are mathematical operations that answer the question: "To what power must a certain base be raised, to obtain a given number?" This is especially useful in reversing exponential operations.
Logarithm properties allow us to simplify complex expressions by leveraging rules such as:

• Product Rule: \( \ln(ab) = \ln a + \ln b \)
• Quotient Rule: \( \ln \left( \frac{a}{b} \right) = \ln a - \ln b \)
• Power Rule: \( \ln(a^b) = b \ln a \)

In the given exercise, for part (a), the Power Rule is crucial. It helps in transforming the expression \(-2 \ln 5\) into \(\ln (5^{-2})\). Similarly, for part (b), understanding that \(\ln e^{2n} = 2n\) involves acknowledging \(\ln e^x = x\). These properties form the backbone of simplifying logarithmic expressions effectively.

Exponential Properties

Exponential functions, where the variable is the exponent, are inherently related to logarithms. They have their unique set of properties that allow us to manipulate and simplify expressions efficiently.

• Inverse Relationships: The expression \(e^{\ln x} = x\) demonstrates that exponentials and logarithms are inverse functions, meaning they nullify each other when used together.
• Zero Exponent Rule: Any number raised to the power of zero equals one, \(a^0 = 1\).
• Negative Exponent: \(a^{-n} = \frac{1}{a^n}\) transforms negative exponents into fractions.

In the problem, part (a) uses the inverse relationship \(e^{\ln x} = x\) to transform \(e^{\ln(5^{-2})}\) directly into \(5^{-2}\). This guides us to then apply the negative exponent rule, simplifying \(5^{-2}\) to \(\frac{1}{25}\). Understanding these properties is essential to breaking down and solving exponential expressions.

Simplifying Expressions

Simplifying expressions is about reducing complex mathematical expressions into their most basic form. It involves recognizing patterns and applying mathematical rules consistently.
The main aim is to make the expression easier to interpret and solve without a calculator.

• Recognize Inverses: Understanding how logarithms and exponents cancel each other out, like \(e^{\ln x} = x\), helps streamline many complex expressions.
• Combine Like Terms: Use properties like the power rule and inverse rules to combine terms and reduce them to simpler equivalents.
• Express in Simple Fractions: For negative exponents, convert them into fractions to simplify results, such as changing \(5^{-2}\) to \(\frac{1}{25}\).

In the exercise, the solution for part (a) precisely demonstrated these concepts. By identifying \(-2 \ln 5\) as a candidate for transformation via logarithmic rules and converting to its simplest fraction form, the expression was successfully simplified to a value of \(\frac{1}{25}\). This approach ensures clarity and conciseness, making algebraic expressions manageable and understandable.