Problem 41 - Calculus Early Transcendentals

Question

Determine whether the following statements are true and give an explanation or counterexample. Assume \(x>0\) and \(y>0\) a. \(\ln x y=\ln x+\ln y\) b. \(\ln 0=1\) c. \(\ln (x+y)=\ln x+\ln y\) d. \(2^{x}=e^{2 \ln x}\)

Step-by-Step Solution

Answer

Explain your answer. Answer: No, the statement \(\ln (x+y) = \ln x + \ln y\) is not always true. Unlike the product rule for logarithms, there is no general rule for the logarithm of a sum. A counterexample is when \(x = 1\) and \(y = 1\), in which case \(\ln 2 \neq 0\).

Step 1: a. Is \(\ln xy = \ln x + \ln y\) true?

Yes, this statement is true. This property is called the product rule for logarithms. We can prove this using the definition of a logarithm. Let's denote \(z = \ln x\) and \(w = \ln y\). Then, by definition, we have \(x = e^z\) and \(y = e^w\). Now, we need to determine \(\ln(xy)\). \(xy = (e^z)(e^w) = e^{z+w}.\) Thus, \(\ln(xy) = \ln(e^{z+w}) = z+w = \ln x + \ln y\). So, the statement is true.

Step 2: b. Is \(\ln 0 = 1\) true?

No, this statement is false. The natural logarithm, \(\ln x\), is only defined for \(x > 0\). Since \(0\) is not greater than \(0\), \(\ln 0\) is undefined. A counterexample or explanation is not necessary in this case, as the statement itself is not well-defined.

Step 3: c. Is \(\ln (x+y) = \ln x + \ln y\) true?

No, this statement is false. Unlike the product rule for logarithms, there is no general rule for the logarithm of a sum. To provide a counterexample, let's use specific values for \(x\) and \(y\). Let \(x = 1\) and \(y = 1\). Then, we have: \(\ln (1+1) = \ln 2 \approx 0.6931\) \(\ln 1 + \ln 1 = 0 + 0 = 0\) Since \(\ln 2 \neq 0\), the statement is false.

Step 4: d. Is \(2^x = e^{2\ln x}\) true?

Yes, this statement is true. We can prove this by utilizing properties of logarithms and exponents. Given the statement \(2^x = e^{2\ln x}\), let's rewrite the base \(2\) as an exponential with base \(e\) using the logarithm definition. Recall that \(2=e^{\ln2}\). Now we have: \((e^{\ln 2})^x = e^{2\ln x}\). Simplify this further: \(e^{x\ln 2} = e^{2\ln x}\). Since the exponential function is injective, we can conclude that \(x\ln 2 = 2\ln x\). Hence, the statement is true.

Key Concepts

Product Rule for Logarithms

The Product Rule for Logarithms is a fundamental property that helps simplify logarithmic expressions involving multiplication. This rule states that the logarithm of a product is equal to the sum of the logarithms of each factor. Mathematically, this can be expressed as follows:

\[ \ln(xy) = \ln x + \ln y \]
Let's delve deeper into why this rule works. Logarithms allow us to "break down" the operation of multiplication into simpler addition, which is generally easier to handle. To understand this property, consider when you have two numbers, say \(x\) and \(y\), both greater than zero. According to the product rule, by taking the natural logarithm of their product \((xy)\), it's the same as adding together the logarithms of \(x\) and \(y\) separately.

• Suppose \( \ln x = a \) and \( \ln y = b \). So by definition, \( x = e^a \) and \( y = e^b \).
• Multiply these expressions to find their product: \( xy = e^a \cdot e^b = e^{a+b} \)
• Taking the logarithm again simplifies to: \( \ln(xy) = \ln(e^{a+b}) = a + b \).

This breakdown explains why \( \ln x + \ln y = \ln (xy) \). This rule is particularly useful in simplifying the complexities involved in logarithmic equations and is widely applied throughout mathematics and science.

Natural Logarithm

A Natural Logarithm is a specific logarithmic function denoted as \( \ln \), which has the constant \( e \) (approximately 2.71828) as its base. When we refer to \( \ln x \), we're asking "what power do we need to raise \( e \) to get \( x \)?"

The natural logarithm has some special properties:

• It's only defined for positive numbers \( x > 0 \).
• \( \ln 1 = 0 \), because \( e^0 = 1 \).
• \( \ln e = 1 \), since \( e^1 = e \).

This particular logarithm is crucial for calculations involving growth processes in disciplines such as biology, economics, and physics, given \( e \)'s natural occurrence in many natural and social phenomena.
In the exercise, the understanding of natural logarithms clarifies why some statements like \( \ln 0 = 1 \) are not true. Since \( \ln x \) cannot be defined for nonpositive values, claiming \( \ln 0 \) equals any real number defies the very definition of the natural logarithm. Instead, approaching \( x = 0 \) from the positive side, the function heads towards negative infinity.

Exponential Functions

Exponential Functions are mathematical expressions in which a constant base is raised to a variable exponent. They are one of the most important types of functions in mathematics, expressed in the form \( f(x) = a^x \), where \( a \) is a positive constant and \( x \) is the exponent.

One well-known exponential function involves the base \( e \), also known as the "natural exponential function," denoted by \( f(x) = e^x \). Exponential functions show up frequently in real-world phenomena:

• Compounded interest in finance.
• Population growth models in biology.
• Radioactive decay in physics.

They grow or decay at a rate proportional to their current value, making them suited for modeling continuous processes.
In the given exercise, exponential functions help verify the truth of statements like \( 2^x = e^{2\ln x} \). Understanding these functions and their linkage to logarithms, such as using properties of \( e \), enables simplification through transformation of bases or applying logarithmic identities. Mastery of manipulating exponential and logarithmic expressions is crucial for tackling sophisticated problems in calculus and beyond.