Problem 18 - Precalculus Mathematics for Calculus

Question

Evaluate the expression. $$\ln \left(\ln e^{-x}\right)$$

Step-by-Step Solution

Answer

The expression is undefined for non-negative \(x\); it's valid only if \(x < 0\).

Step 1: Simplify the Inner Expression

First, simplify the expression inside the innermost logarithm: \[ e^{-x} \]The base \(e\) exponentiated to \(-x\), gives us \(e^{-x}\).

Step 2: Evaluate the Natural Logarithm

Now apply the natural logarithm to \(e^{-x}\): \[ \ln(e^{-x}) = -x \]This results from the property of logarithms: \(\ln(a^b) = b \cdot \ln(a)\), where \(\ln(e) = 1\).

Step 3: Evaluate the Outer Natural Logarithm

Finally, apply the natural logarithm to the result from Step 2:\[ \ln(-x) \]However, the natural logarithm \(\ln(c)\) is only defined for positive numbers \(c\). Therefore, \(\ln(-x)\) is only valid if \(-x > 0\), which implies that \(x < 0\). Thus, the expression is not defined for non-negative \(x\).

Key Concepts

Natural Logarithm

The natural logarithm, often denoted as \(\ln\), is a special type of logarithm that has the base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828. It is widely used in mathematics, especially in calculus, due to its unique properties.
The natural logarithm of a number \(a\), represented as \(\ln(a)\), answers the question: what power do we raise \(e\) to, in order to get \(a\)? For instance, \(\ln(e) = 1\) because \(e^1 = e\).
When you evaluate expressions involving the natural logarithm, it is crucial to remember the importance of its base. The closeness of \(\ln\) to calculus operations like differentiation and integration allows for simplifications. For example, the derivative of \(\ln(x)\) is \(1/x\), and the integral of \(1/x\) results into \(\ln|x| + C\), where \(C\) is the constant of integration.

• The natural logarithm simplifies certain calculations, making it a favorite among mathematicians.
• In logarithmic equations, understanding \(\ln(a)\) can help in solving for unknowns with exponential growth or decay, such as in the expression \(\ln(e^{-x})\) yielding \(-x\).

Properties of Logarithms

The properties of logarithms are powerful tools that help simplify and solve logarithmic expressions. These properties are derived from the fundamental concept of logarithms as the inverse of exponential functions.

Here are some key properties:

• Product Property: \(\log_b(mn) = \log_b(m) + \log_b(n)\). This property states that the logarithm of a product is the sum of the logarithms.
• Quotient Property: \(\log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n)\). This indicates that the logarithm of a quotient is the difference of the logarithms.
• Power Property: \(\log_b(m^n) = n \cdot \log_b(m)\). For the natural logarithm, this acts as \(\ln(a^b) = b \cdot \ln(a)\).
• Change of Base Formula: \(\log_b(a) = \frac{\log_k(a)}{\log_k(b)}\), useful for converting logarithms to a convenient base.

The step-by-step solution provided in the original exercise utilized the power property, allowing us to simplify \(\ln(e^{-x})\) to \(-x\). Understanding and applying these properties can simplify complex expressions and solve logarithmic equations efficiently.

Domain of Logarithm Functions

The domain of a logarithmic function is an essential aspect to consider when dealing with such functions. The domain refers to all the possible input values (or arguments) of the function that lead to real-number outputs. For logarithmic functions, these inputs must be positive.

For example, the natural logarithm function \(\ln(x)\) is only defined for values of \(x\) that are greater than 0. This implies that you cannot take the natural logarithm of a negative number or zero, making the domain \((0, \infty)\).
This restriction stems from the definition of a logarithm as the inverse of an exponential function. Since exponential functions yield only positive outputs for real inputs, their inverses, logarithms, can only accept positive inputs.
Applying this understanding to the exercise, we recognize that \(\ln(-x)\) is defined only if \(-x\) is positive, or simply when \(x < 0\). Such insight not only aids in solving logarithmic expressions but is also crucial for avoiding errors in domain-related mathematical problems.