Problem 63 - Elementary and Intermediate Algebra

Question

Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ \ln x=1 $$

Step-by-Step Solution

Answer

The exact solution is \( x = e \), and the approximation is \( x \approx 2.7183 \).

Step 1: Identify the Equation Type

The given equation is \( \ln x = 1 \), which is a logarithmic equation, specifically involving the natural logarithm, \( \ln(x) \). We need to find the value of \( x \) satisfying this equation.

Step 2: Express in Exponential Form

Recall that the natural logarithm \( \ln(x) \) is equivalent to \( \log_e(x) \), where \( e \) is the base of the natural logarithm. Therefore, to solve \( \ln x = 1 \), we convert to exponential form: \( x = e^1 \).

Step 3: Calculate the Exact Solution

The expression \( e^1 \) simplifies to \( e \), so \( x = e \) is the exact solution to the equation \( \ln x = 1 \).

Step 4: Calculate the Approximation

Using a calculator, we find that \( e \) is approximately 2.718281828. To four decimal places, this approximates to \( x \approx 2.7183 \).

Key Concepts

Understanding Logarithmic Equations

A logarithmic equation is one where the variable you want to solve for is inside of a logarithm. In our equation, \( \ln x = 1 \), we're dealing with a natural logarithm. The natural logarithm, represented by \( \ln \), uses the base \( e \), an important mathematical constant approximately equal to 2.718. In this equation, the objective is to unravel \( x \) from the log function to find its exact value. When you have an equation set in this form, it is often helpful to think about converting it into an exponential equation, which can make it much simpler to solve. By understanding that logs and exponential equations are inverse operations, we can easily maneuver between them to isolate the variable.

Exponential Form and Its Role

The exponential form of an equation is a way to express the relationship between numbers with a given base raised to a power. In the context of our specific problem, we start with the equation \( \ln x = 1 \), where the task is to solve for \( x \).

This can be converted from a logarithmic to an exponential form using the understanding that \( \ln(x) = \log_e(x) \). The exponential form is essentially saying that \( x \) is equal to \( e \) raised to the power of the other side of the equation, which leads us to \( x = e^1 \).

Converting to this form helps unravel \( x \) so that we can find its exact solution. In many logarithmic equations, identifying and converting to an exponential form is a critical skill because it permits us to isolate the unknown variable more easily.

Approximating Values

Once we find the exact solution of an equation, it is often beneficial or required to provide an approximation. Approximations simplify complex numbers to be more manageable in practical scenarios, especially when dealing with irrational numbers like \( e \).

In our problem, after converting to exponential form, we found \( x = e^1 \), which deduces down to \( x = e \). We used a calculator to approximate \( e \), which is about 2.718281828. However, we exact the approximation to four decimal places, providing \( x \approx 2.7183 \).

Providing approximations, especially to four decimal places or more, is often essential in scientific and engineering tasks, where precision is crucial. It allows us to present a value that is easier to use in other calculations while understanding that it is an estimate based on the true value.