Question

Explain why or why not 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. \(2^{x}=e^{2 \ln x}\) e. The area under the curve \(y=1 / x\) and the \(x\) -axis on the interval \([1, e]\) is 1

Short answer

Question: Determine the truth of the following statements and provide explanations or counterexamples. a) \(\ln xy = \ln x + \ln y\) b) \(\ln 0 = 1\) c) \(\ln(x+y) = \ln x + \ln y\) d) \(2^x = e^{2\ln x}\) e) The area under the curve \(y=\frac{1}{x}\) and the x-axis on the interval \([1, e]\) is 1. Answer: a) True b) False c) False d) True e) True

Step-by-step solution

Statement a: \(\ln xy = \ln x + \ln y\)

This statement is true. According to the logarithm properties, the product rule states that \(\log_b (mn) = \log_b m + \log_b n\). Thus, for natural logarithms (with base \(e\)), it holds true that: $$\ln(xy) = \ln(x) + \ln(y)$$

Statement b: \(\ln 0 = 1\)

This statement is false. The natural logarithm is not defined for non-positive arguments (x ≤ 0). Also, note that if this statement were true, it would mean that \(e^1 = e=0\), which is not true either.

Statement c: \(\ln(x+y) = \ln x + \ln y\)

This statement is false. The given expression does not hold true for all non-negative values of x and y. For instance, let's consider a simple counterexample: Let \(x = 1\) and \(y=1\). Then, substituting the values in the given expression: $$\ln(1+1) = \ln 1 + \ln 1$$ $$\ln 2 = 0 + 0$$ However, \(\ln 2\) is not equal to 0, so the statement is false.

Statement d: \(2^x = e^{2\ln x}\)

This statement is true. We can rewrite \(2^x\) using \(e\) as its base, because: $$2^x = e^{\ln(2^x)}$$ Now, using the property of logarithm \(log_b (m^n) = n\log_b m\), we have: $$e^{\ln(2^x)} = e^{x\ln 2}$$ On the other side, using the logarithm property \(\log_b (mn) = \log_b m + \log_b n\), we have: $$e^{2\ln x} = e^{\ln x + \ln x} = e^{\ln x^2}$$ Now, since \(2^x = e^{x\ln 2}\) and \(e^{x\ln 2} = e^{\ln x^2}\), both the expressions are equal.

Statement e: The area under the curve \(y=\frac{1}{x}\) and the x-axis on the interval \([1, e]\) is 1

To find the area under the curve, we can use definite integration on the interval [1, e]: $$A = \int_{1}^{e} \frac{1}{x} dx$$ The integral of \(\frac{1}{x}\) is \(\ln|x|+C\). Applying the limits, we get: $$A = [\ln|x|]_1^e = \ln e - \ln 1$$ Since \(\ln e = 1\) and \(\ln 1 = 0\), we obtain: $$A = 1 - 0 = 1$$ Hence, the statement is true. The area under the curve \(y=\frac{1}{x}\) and the x-axis on the interval \([1, e]\) is indeed 1.

Key concepts

logarithm properties

Logarithms are powerful mathematical tools that help us simplify multiplication and division operations. One of the key properties of logarithms is the **product rule**. This rule states that the logarithm of a product is the sum of the logarithms of its factors. For natural logarithms, where the base is Euler's number \(e\), the property can be expressed as: \(\ln(xy) = \ln x + \ln y\). This insight allows us to break down complex expressions into simpler, more manageable parts. It's crucial to understand this rule, as it is frequently used in calculus and other advanced mathematics topics.

• Product Rule: \( \ln(xy) = \ln x + \ln y \).
• Common in solving logarithmic equations and modeling scenarios.

Applying this rule correctly is vital for tackling a range of mathematical problems, from simple algebra to more advanced calculus applications.

definite integration

**Definite integration** is a fundamental concept in calculus, representing the net area under a curve. It provides a way to calculate total accumulation within given boundaries. In simple terms, when you perform definite integration, you are summing up tiny slices of area under the curve between a start point and an endpoint on the x-axis.
Consider the integral \( \int_{a}^{b} f(x) \, dx \), which represents the area under the curve \(f(x)\) from \(x=a\) to \(x=b\). The process of definite integration involves calculating the integral of the function within these limits and subtracting the value of the antiderivative evaluated at the lower limit from the value at the upper limit.

• Integral bounds define the section of the curve to analyze.
• The process results in a definite value, unlike indefinite integration.

This concept is particularly useful in areas where determining precise quantities, such as total distance traveled or total profit generated, is necessary.

area under the curve

The **area under the curve** is a way to measure the space between a curve and the x-axis, within specified limits. In mathematical terms, it refers to the integral of a function over an interval. The idea of finding the area under the curve is crucial in many practical applications, such as calculating accumulated quantities and understanding the overall behavior of functions.
In our exercise, to find the area under the curve \( y = \frac{1}{x} \) from \( x = 1 \) to \( x = e \,\), we used definite integration: \( \int_{1}^{e} \frac{1}{x} \, dx \.\). The solution correctly identified this area as 1, using properties of natural logarithms.

• Used to calculate real-world quantities like displacement or economic gain.
• Integral calculus provides the tools to measure these areas accurately.

Mastering this concept allows for deeper insights into continuous data and processes in diverse fields like physics and economics.

product rule

The **product rule** is a concept often encountered in calculus when dealing with derivatives, but it is also relevant in logarithms as discussed earlier. Specifically, in differentiation, the product rule provides a way to differentiate the product of two functions. If you have two functions, say \(f(x)\) and \(g(x)\), their derivative can be found using the product rule: \( (f \cdot g)' = f' \cdot g + f \cdot g' \). This means you take the derivative of the first function multiplied by the second, plus the first function times the derivative of the second.
This rule is integral in calculus as it allows you to solve problems involving complex functions created by multiplying simpler functions.

• Applies to both natural logarithms and calculus.
• Essential for solving higher-order derivative problems.

Understanding the product rule is critical for efficiently tackling calculus problems where functions are multiplied.