Question

The following exercises are intended to derive the fundamental properties of the natural log starting from the definition ln(x)=?x1dtt, using properties of the definite integral and making no further assumptions. Use a change of variable in the integral \(\int_{1}^{x y} \frac{1}{t} d t\) to show that \(\ln x y=\ln x+\ln y\) for \(x, y>0\).

Short answer

Thus, \( \ln(xy) = \ln(x) + \ln(y) \) for \(x, y > 0\).

Step-by-step solution

Write the Definition of Natural Logarithm

The natural logarithm of a number \(x\) is defined using an integral: \( \ln(x) = \int_{1}^{x} \frac{1}{t} \, dt \). Start by considering the given integral \( \ln(xy) = \int_{1}^{xy} \frac{1}{t} \, dt \).

Use Substitution in the Integral

To find \( \ln(xy) \), perform a change of variable by letting \( u = \frac{t}{y} \). Then, \( t = uy \) and \( dt = y \, du \). Adjust the integral limits: when \( t = 1 \), \( u = \frac{1}{y} \); when \( t = xy \), \( u = x \). Thus, the integral becomes \( \int_{1/y}^{x} \frac{1}{uy} \, y \, du = \int_{1/y}^{x} \frac{1}{u} \, du \).

Break Down the Integral into Two Parts

Split the integral into two separate integrals by recognizing that you can express it from 1 to x and from 1/y to 1. This yields \( \int_{1/y}^1 \frac{1}{u} \, du + \int_1^x \frac{1}{u} \, du \). Based on the definition of the natural logarithm, the first integral becomes \(-\ln(y)\) and the second integral becomes \(\ln(x)\).

Combine the Integrals

Combine the results of the integrals: the expression becomes \( -\ln(y) + \ln(x) = \ln(x) - \ln(y) \). Notice that this is equivalent to rearranging the original problem \( \ln(xy) = \ln(x) + \ln(y) \) based on properties of logarithms.

Key concepts

Definite Integral

A definite integral is a concept that provides a way to calculate the accumulation of a quantity, like area under a curve, between two specific points. When we talk about the integral \(\int_{a}^{b} f(t) \, dt\), it gives us the total amount of the function \(f(t)\) accumulated between the limits \(a\) and \(b\).
In the context of the natural logarithm, we use the definite integral to define \(\ln(x)\), where \(x > 0\). The integral starts from 1 and goes to \(x\):

• \( \ln(x) = \int_{1}^{x} \frac{1}{t} \, dt \)

This specific integral calculates the area under the curve of \(\frac{1}{t}\) from 1 to \(x\), which directly ties into the fundamental definition of the natural logarithm. Understanding definite integrals is crucial because they act as a bridge connecting the abstract mathematical function to practical numerical values.
This is especially pertinent in problems involving natural logarithms, where properties often rely on recognizing how these integrals behave between different limits.

Change of Variable

The change of variable is a crucial technique in calculus used to simplify integrals. By substituting a new variable into an integral, we transform it into a more manageable form. Take the original integral \( \int_{1}^{xy} \frac{1}{t} \, dt \). Performing a change of variable helps evaluate this integral to prove properties of logarithms.
When we let \( u = \frac{t}{y} \), it simplifies the process. This changes the bounds and the expression inside the integral:

• The variable transformation means that when \( t = 1 \), \( u = \frac{1}{y} \), and when \(t = xy\), \( u = x\).
• This conversion offers new integral limits from \( 1/y \) to \( x \) and replaces the integrand \( \frac{1}{t} \) with \( \frac{1}{u} \).
• It leads to \( \int_{1/y}^{x} \frac{1}{u} \, du \).

This approach shows how changing variables smartly can simplify integration tasks, particularly when proving fundamental equations like those involving logarithms.

Logarithm Properties

Logarithm properties are foundational tools in mathematics that help us manipulate and simplify expressions involving logarithms. One such property is \( \ln(xy) = \ln(x) + \ln(y) \).
This reflects the way logarithms relate multiplication into addition, which is why they're so powerful in solving otherwise complex mathematical problems.Using definite integration, when we evaluate the integrals \( \int_{1}^{1/y} \frac{1}{u} \, du \) and \( \int_{1}^{x} \frac{1}{u} \, du \), their results are:

• \( -\ln(y) \)
• \( \ln(x) \)

Combining them shows \( -\ln(y) + \ln(x) = \ln(x) - \ln(y) \), which can be rearranged to showcase that multiplicative relationships in arguments of logarithms become additive relationships.This property of logarithms is pivotal in calculations and transformations in mathematics, providing simpler ways to work through otherwise cumbersome multiplication issues.

Substitution Method

The substitution method is a technique often used in integration to simplify an integral, making it easier to handle by converting it into a familiar form. In our example problem, we employed the substitution method by setting \( u = \frac{t}{y} \).
This changed the integral in helpful ways:

• With \( t = uy \) and \( \frac{dt}{du} = y \), we replaced the integrand and adjusted the limits accordingly.
• This allowed direct computation with respect to \( u \), simplifying the original integration task.
• It created integral pieces that directly apply known results about logarithms.

This methodology streamlines complex integration problems by transforming them into known configurations, thereby eliciting straightforward results that either follow directly from definitions or recognized properties like those of logarithms.
Effectively using substitution can substantially simplify an integral, enhancing our ability to leverage powerful mathematical tools like logarithm properties in proofs and computations.