Problem 97 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, determine whether the statement is true or false given that \(f(x)=\ln x .\) If it is false, explain why or give an example that shows it is false. $$ f(x-2)=f(x)-f(2), \quad x>2 $$

Step-by-Step Solution

Answer

The given statement \(f(x-2)=f(x)-f(2)\) is false for the function \(f(x) = \ln x\). This is because when the values are substituted and simplified in the equation, it results in a false statement.

Step 1: Understand the Function

The function here is \(f(x) = \ln x\). The logarithm function has a property that states \(\ln a - \ln b = \ln \left(a/b\right)\). This critical property of logarithms will aid in solving this exercise.

Step 2: Substitute the Values

Let's substitute the values of \(x-2\) and \(x\) in the function \(f(x)=\ln x\). So, the statement \(f(x-2)=f(x)-f(2)\) becomes \(\ln (x-2)= \ln x - \ln 2\).

Step 3: Simplify the Equation

Using the logarithm property \(\ln a - \ln b = \ln \left(a/b\right)\), we can simplify \(\ln x - \ln 2\) to \(\ln \left(x/2\right)\). So, our equation becomes \(\ln (x-2)= \ln \left(x/2\right)\).

Step 4: Analyze the Equation

here comparing both sides, it's clear that \(x - 2\) does not equal to \(x/2\) for values \(x > 2\). Thus, the given statement is false. An example would be \(x = 3\): substituting this into both sides results in \(\ln 1\) on the left side and \(\ln 1.5\) on the right side, which are not equal.

Key Concepts

Properties of Logarithms

Logarithmic functions possess unique properties that simplify the handling of logarithmic expressions. One fundamental property to remember is \[ \ln a - \ln b = \ln \left(\frac{a}{b}\right) \].
This property is essential when working with logarithmic equations, as it allows you to combine or break down expressions into simpler forms.
Let's consider an example: if you have \( \ln 10 - \ln 2 \), you can simplify it using this property to get \( \ln \left(\frac{10}{2}\right) = \ln 5 \).
This simplification makes it easier to solve equations and analyze functions.Remember this key point:

• Subtraction of logarithms translates into the logarithm of a division.
• It is a crucial tool for simplifying expressions involving logarithms.

Understanding these properties makes logarithms less intimidating and more manageable for solving various math problems.

Function Analysis

Let's delve into function analysis with our specific function, \( f(x) = \ln x \). Analyzing a logarithmic function involves understanding its domain, behavior, and transformations.
Firstly, it's important to note that the natural logarithm \( \ln x \) is only defined for \( x > 0 \). This means \( f(x) \) doesn't exist for non-positive numbers.
In the context of \( x - 2 \), \( x \) must be greater than 2 for the function to remain in its domain.Moreover:

• Logarithmic functions like \( \ln x \) are continuously increasing. This means as \( x \) grows, \( f(x) \) also grows.
• The graph of \( \ln x \) is always rising but does so more slowly as \( x \) becomes larger, known as a decreasing rate of increase.

Analyzing the behavior of a function gives insights into how it reacts under various manipulations, helping us identify true or false equations.

Equation Simplification

When faced with complex equations, simplifying them can make the problem-solving process significantly easier. Simplification involves using properties and substituting values to obtain clearer expressions.
In the original exercise, we started with the function \(f(x) = \ln x\) and needed to simplify \(f(x-2) = f(x) - f(2)\).
By substituting, we reached \( \ln (x-2) = \ln x - \ln 2 \).Simplification steps included:

• Recognizing this as a classic form suitable for the property \( \ln a - \ln b = \ln \left(\frac{a}{b}\right) \).
• Using the property to transform \( \ln x - \ln 2 \) into \( \ln \left(\frac{x}{2}\right) \).

These steps help transform cumbersome expressions into manageable ones.
The resulting equation \( \ln (x-2) = \ln \left(\frac{x}{2}\right) \) highlights discrepancies, leading to the conclusion that the statement is false.