Problem 83 - Precalculus with Calculus Previews

Question

The following question appeared on an examination: Find the domain of the function \(f(x)=\ln \left(\frac{x-3}{x}\right)\). One student reasoned that using the laws of logarithms the function \(f\) could be rewritten as $$ f(x)=\ln (x-3)-\ln x $$ Because the domain of \(\ln (x-3)\) is the interval (3, \infty) and the domain of \(\ln x\) is the interval \((0, \infty)\), the domain of \(f\) is the intersection \((0, \infty) \cap(3, \infty)\) \(=(3, \infty)\). Discuss: Is the student's reasoning valid?

Step-by-Step Solution

Answer

The student's reasoning is valid; the domain is \((3, \infty)\).

Step 1 - Understanding the Logarithmic Expression

The function given is \( f(x) = \ln \left( \frac{x-3}{x} \right) \). Recall that the domain of the logarithm \( \ln(a) \) is \( a > 0 \). Thus, for \( \ln \left( \frac{x-3}{x} \right) \), we need \( \frac{x-3}{x} > 0 \) to determine the domain.

Step 2 - Set Up the Inequality for the Logarithm

We need to solve \( \frac{x-3}{x} > 0 \). This inequality can be interpreted as when the expression is positive. This occurs when the numerator and the denominator have the same sign (both positive or both negative).

Step 3 - Analyze Numerator and Denominator

The numerator \( x-3 > 0 \) implies \( x > 3 \), and the denominator \( x > 0 \) defines \( x > 0 \). Both must be combined under the condition that the fraction itself is positive.

Step 4 - Solve the Combined Inequality

For \( \frac{x-3}{x} > 0 \) to hold:- The product \((x-3)(x) \) should be positive.- Solve two cases: 1. \( x > 3 \) and \( x > 0 \), which is satisfied by \( x > 3 \). 2. \( x < 3 \) and \( x < 0 \), which is invalid for real numbers since no \( x \) satisfies both simultaneously.Therefore, \( x > 3 \) is the solution.

Step 5 - Compare with Student's Reasoning

The student found the domain as \( (3, \infty) \) by taking the intersection of the domains of \( \ln(x-3) \) and \( \ln(x) \), ignoring the requirement for the argument of the logarithm to be positive. The correct domain of the function is \( (3, \infty) \) because that ensures \( \frac{x-3}{x} > 0 \), agreeing with what we found above but through an incorrect assumption the student made.

Key Concepts

Logarithmic Functions

Logarithmic functions are mathematical expressions that are the inverse of exponential functions. Specifically, for a logarithm \(\ln x\), it represents the power to which the base 'e' (approximately 2.718) must be raised to obtain the number 'x'. One key aspect of logarithmic functions is their domain, which includes only positive numbers.
This means that when you see \(\ln(a)\), the value inside the logarithm \(a\) must be greater than zero. This is crucial because if \(a\) were zero or negative, the logarithmic function would not be defined. In solving domain problems, especially when involving logarithmic expressions like \(\ln\left(\frac{x-3}{x}\right)\), it must be ensured that the expression inside the log function remains positive.
When solving for the domain, you focus on ensuring that the entire expression inside the logarithm stays positive, which determines the values for which the function is defined.

Inequalities

Understanding inequalities is fundamental when working with domains in mathematical functions. An inequality expresses the relation between two expressions that are not equal, using symbols such as \(>\), \(<\), \(\geq\), and \(\leq\). For functions like \(\ln\left(\frac{x-3}{x}\right)\), we use inequalities to determine where the expression \(\frac{x-3}{x}\) is greater than zero.

To solve \(\frac{x-3}{x} > 0\), both the numerator and denominator must have the same sign. That is, both should be positive, or both should be negative.

• If both are positive: Solve \(x - 3 > 0\) and \(x > 0\).
• If both are negative: Solve \(x - 3 < 0\) and \(x < 0\), though this results in no valid real numbers for this specific scenario.

Combining both computations gives the solution \(x > 3\), which indicates that the fraction \(\frac{x-3}{x}\) is positive, conforming with the function’s logarithmic requirement.
Inequalities play a vital role in determining valid values for mathematical expressions, especially in ensuring the domain of a function is correct.

Function Composition

Function composition involves combining two or more functions to form a new function. In this process, the output of one function becomes the input of another. It is a powerful tool in mathematics, allowing functions to be simplified and combined.

In the domain context, composing functions can sometimes further restrict the domain, as each function has its own domain restrictions. When evaluating \(f(x) = \ln(x-3) - \ln(x)\) as in the student’s reasoning, function composition hints at the interactions between these sub-functions.

• The domain of \(\ln(x-3)\) is \(x > 3\) because \(x-3\) must be positive.
• The domain of \(\ln(x)\) is \(x > 0\) since \(x\) is inside a logarithm.

Thus, you must consider how all these constraints combine. The intersection of the domains \((0, \infty) \cap (3, \infty)\) results in \((3, \infty)\). This matches, coincidentally, the calculated domain, showing how overlapping domain conditions work in compounded functions. The key takeaway in function composition is considering each function's domain to determine the resulting function's domain accurately.