Question

In Exercises, use a graphing utility to verify that the functions are equivalent for \(x>0\). $$ \begin{aligned} &f(x)=\ln \frac{x^{2}}{4} \\ &g(x)=2 \ln x-\ln 4 \end{aligned} $$

Short answer

Yes, the functions \(f(x)=\ln \frac{x^{2}}{4}\) and \(g(x)=2 \ln x - \ln 4\) are equivalent for \(x>0\). For any given \(x\) greater than 0, both functions give the same output, as per the rules of logarithms.

Step-by-step solution

Express Functions

First, we will express the given functions: The first function \(f(x)=\ln \frac{x^{2}}{4}\) and the second function \(g(x)=2 \ln x - \ln 4\). We need to confirm that these two functions are alike for values of \(x > 0\).

Simplify the First Function

To confirm that \(f(x)\) is the same as \(g(x)\), let's begin by simplifying the first function. We know the rules of logarithms, specifically the quotient rule which states that: \(\ln (a/b) = \ln a - \ln b\). Applying this to our function, we get \(f(x) = \ln x^2 - \ln 4\).

Apply the Exponent Rule of Logarithms to f(x)

The exponent rule states that \(\ln a^n = n \ln a\). Applying this to our first term for \(f(x)\) in the previous step , we get \(f(x) = 2\ln x - \ln 4\).

Compare the Functions

Now, when we look at the second function, \(g(x) = 2 \ln x - \ln 4\), we see that it is identical to our modified first function, which is \(f(x) = 2\ln x - \ln 4\). So, we can say that \(f(x)\) and \(g(x)\) are equivalent.

Key concepts

Logarithm Rules

Logarithm rules are essential tools for simplifying complex expressions like those found in this exercise. Let’s explore the key rules:

• Product Rule: \[\ln(a \cdot b) = \ln a + \ln b\] This rule helps when multiplying numbers inside a logarithm.
• Quotient Rule: \[\ln\left(\frac{a}{b}\right) = \ln a - \ln b\] Using this, we can break down a fraction into two simpler logarithms.
• Exponent Rule: \[\ln(a^n) = n \ln a\] This allows you to move an exponent in front of the logarithm for easier handling.

In the given exercise, these rules were applied to simplify \(f(x) = \ln\left(\frac{x^2}{4}\right)\) into \(f(x) = 2 \ln x - \ln 4\). By using the quotient rule, the fraction turned into a subtraction, and the exponent rule transformed \(ln(x^2)\) into \(2 \ln x\).

Understanding these principles makes working with logarithms much more manageable.

Function Equivalence

Function equivalence occurs when two functions yield the same output for every input in their domain. In this exercise, we are tasked with proving that \(f(x)\) and \(g(x)\) are equivalent when \(x > 0\).

Here’s why function equivalence is crucial:

• It confirms that two seemingly different expressions actually represent the same relationship, which is a powerful simplification tool.
• Ensures that transformations and simplifications in mathematics don't change the original problem's meaning.

Simplifying \(f(x)\) using logarithm rules allowed us to transform it into \(g(x) = 2\ln x - \ln 4\). Since this equals our expression for \(f(x)\), we confirmed their equivalence. Equivalence is particularly useful in calculus and algebra, where complex problems are often expressed in different, yet equivalent, forms for simplicity or computational efficiency.

Graphing Utilities

Graphing utilities, such as graphing calculators or software, are valuable tools in visualizing mathematical functions. They help in verifying equivalence and understanding the behavior of functions.

Why use graphing utilities when comparing functions?

• They provide immediate visual evidence of function equivalence by showing overlapping graphs.
• Help explore behavior, such as asymptotes and intercepts, that algebraic methods might not immediately reveal.
• Offer dynamic manipulation, letting you adjust values to see real-time changes.

In the exercise, using a graphing utility would show that the graphs of \(f(x) = \ln\left(\frac{x^2}{4}\right)\) and \(g(x) = 2\ln x - \ln 4\) overlap completely for \(x > 0\). This visual validation supports the algebraic proof of equivalence and deepens understanding of how such functions behave across different domains.