Problem 93 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the logarithmic equation algebraically. Round the result to three decimal places. Verify your answer(s) using a graphing utility. $$\ln x=-3$$

Step-by-Step Solution

Answer

The solution to the logarithmic equation \(\ln x = -3\) is \(x ≈ 0.050\).

Step 1: Rewrite the Equation

The given equation is \(\ln x = -3\). Let's rewrite it in exponential form. In general, the natural logarithm \(\ln a = b\) is equivalent to the exponential equation \(e^b = a\). So here, the equation becomes \(e^{-3} = x\).

Step 2: Compute the Value of \(x\)

After rewriting the equation, we need to calculate \(e^{-3}\). Using a calculator, \(e^{-3} ≈ 0.050\). So, \(x ≈ 0.050\).

Step 3: Verify the solution

Substitute \(x\) back into the original equation to validate. We get \(\ln 0.050\), which should come out to be \(-3\). This verifies that the calculated value of \(x\) is correct.

Key Concepts

Understanding the Natural Logarithm

The natural logarithm, denoted as \( ln \) and spoken as 'elog n,' is a logarithmic function with a base of \( e \), the Euler's number, approximately equal to 2.71828. It's a critical concept in algebra and calculus, with various applications in science and engineering. When you encounter an equation like \( ln x = -3 \), it essentially represents the power to which we must raise \( e \) to obtain \( x \).

Transforming to Exponential Form

Exponential form is essential in solving logarithmic equations because it allows us to directly evaluate the variable. If \( ln a = b \), then the corresponding exponential form is \( e^b = a \). In the case of the given problem, we rewrite \( ln x = -3 \) into \( e^{-3} = x \).

This transformation simplifies the solving process, providing a more straightforward calculation, which is especially useful for equations that do not yield integers or easily recognizable fractional solutions. It's important to be comfortable with both the logarithmic and exponential forms, as they are inversely related operations.

Converting to Exponential Form for Solution

Rewriting logarithmic equations in exponential form can reveal solutions that are not readily apparent. After transforming \( ln x = -3 \) into \( e^{-3} = x \) using the property \( ln a = b \) is equivalent to \( e^b = a \) we directly find the value of \( x \) by evaluating \( e^{-3} \).

Using mathematical software or a calculator, one can input \( e^{-3} \) to get the approximate numerical value. For instance, calculating \( e^{-3} \) gives us a value around 0.050. Hence, \( x \) is approximately 0.050, which is the solution to our equation. Although the solution may appear abstract, the process of converting to exponential form and calculating values is vital in solving logarithmic equations efficiently and effectively.

Verification Using a Graphing Utility

One practical way to verify the solution of a logarithmic equation is through the use of a graphing utility. These are tools—such as graphing calculators or graphing software—that allow users to visualize functions and their properties.

How to Use Graphing Utilities

After solving the logarithmic equation algebraically and getting \( x ≈ 0.050 \) as the solution, we can input the original logarithmic expression into a graphing utility. Plotting the function \( y = ln x \) and marking the point \( x = 0.050 \) on the graph can confirm if the value of \( y \) aligns with \( -3 \).

When plotted, the graph should intersect the x-axis at \( x ≈ 0.050 \) if the solution is correct. This serves as a visual confirmation, providing a robust validation of the algebraic solution. Through this process, students gain additional insight into the function's behavior and the accuracy of their solutions. It's a crucial step for ensuring that the algebraic manipulations have led to the correct result, and for understanding the implications of logarithmic functions in a visual format.