Question

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(\ln x-\ln (x+1)=2\)

Short answer

The solution to the logarithmic equation \( \ln x-\ln (x+1)=2 \) is \( x = 2.582 \), approximated to three decimal places.

Step-by-step solution

Use the properties of logarithms

The logarithm difference rule states that \( \ln a - \ln b = \ln \frac{a}{b} \). Therefore, the original equation \(\ln x-\ln (x+1)=2\) can be rewritten as \( \ln \frac{x}{x+1} = 2 \).

Convert the logarithmic equation to an exponential equation

Converting from logarithmic to exponential form helps to simplify the problem further. The conversion can be done as follows: If \( \ln a = b \), then \( a = e^b \). Therefore, the equation \( \ln \frac{x}{x+1} = 2 \) can be written in exponential form as \( \frac{x}{x+1} = e^2 \).

Solve for x

Now \( \frac{x}{x+1} = e^2 \) is a simple algebraic equation. By making the denominator on the left side the same as the denominator on the right side, we can solve for x. Here's how: multiplying both sides by \( x+1 \), we get \( x = e^2(x+1) \). Distributing \( e^2 \) on the right side, we get \( x = e^2x + e^2 \). Setting all terms involving x on one side and numerical terms on the other, we get \( (1-e^2)x = e^2 \). Thus, \( x = \frac{e^2}{1-e^2} \)

Approximate the result

Approximate x to three decimal places using a calculator, and you get \( x = 2.582 \).

Key concepts

Properties of Logarithms

Logarithms have specific properties that simplify solving equations involving them. Let's begin with one of the primary rules: the difference of logarithms. According to this rule,

• \( \ln a - \ln b = \ln \frac{a}{b} \)

This rule allows us to condense the subtraction of two logarithms into a single logarithm of a fraction. In our original problem, \( \ln x - \ln (x+1) = 2 \), we used this property to write the equation as a single logarithmic term:

• \( \ln \frac{x}{x+1} = 2 \)

Another important property of logarithms is that the logarithm of a power can help reduce the complexity of expressions:

• \( \ln a^b = b \cdot \ln a \)

These properties are fundamental tools for transforming and simplifying logarithmic expressions, making them more manageable for solving.

Exponential Equations

Once a logarithmic equation is simplified, converting it to an exponential form can often make it easier to solve. This involves using the inverse relationship between logarithms and exponentials:

• If \( \ln a = b \), then \( a = e^b \)

In our exercise, after applying the properties of logarithms, we arrived at:

• \( \ln \frac{x}{x+1} = 2 \)

Converting this equation, we get:

• \( \frac{x}{x+1} = e^2 \)

Here, \( e \) represents the base of natural logarithms, approximately equal to 2.718. The purpose of converting to the exponential form is to move away from logarithms and work with a form that usually leads to an easier path for finding solutions.

Solving Algebraic Equations

Algebraic equations involve finding the value of a variable that satisfies the equation. Starting from our exponential form:

• \( \frac{x}{x+1} = e^2 \)

The goal is to isolate \( x \). First, multiply both sides by \( x+1 \) to eliminate the fraction:

• \( x = e^2(x+1) \)

Distribute \( e^2 \) on the right-hand side:

• \( x = e^2 x + e^2 \)

Next, gather all terms involving \( x \) on one side:

• \( (1 - e^2)x = e^2 \)

Finally, solve for \( x \) by dividing:

• \( x = \frac{e^2}{1-e^2} \)

This gives the exact solution, but we can further approximate \( x \) to three decimal places using a calculator, yielding \( x \approx 2.582 \). Understanding how to isolate and solve for variables in different types of equations is a core skill in algebra.