Problem 65 - Intermediate Algebra

Question

Solve each equation. Express all answers to four decimal places. $$ \ln x=-3.71 $$

Step-by-Step Solution

Answer

The solution is \( x \approx 0.0246 \).

Step 1: Recall the Definition of a Natural Logarithm

The natural logarithm function, denoted as \( \ln(x) \), is the inverse of the exponential function \( e^x \). This means that if \( y = \ln(x) \), then \( x = e^y \). We will use this property to solve the equation \( \ln x = -3.71 \).

Step 2: Exponentiate Both Sides of the Equation

To isolate \( x \), exponentiate both sides using base \( e \). This transforms the equation from logarithmic to exponential form: \[ e^{\ln x} = e^{-3.71} \]Since \( e^{\ln x} = x \), this simplifies to:\[ x = e^{-3.71} \]

Step 3: Calculate \( e^{-3.71} \) Using a Calculator

Use a scientific calculator to find the value of \( e^{-3.71} \). Entering \( -3.71 \) and then executing the exponential function \\( e^x \) will give the solution. Ensure your calculator is set to the appropriate mode for natural exponentiation. The result is approximately:\( x \approx 0.0245907 \)

Step 4: Round the Result to Four Decimal Places

Round the calculated value of \( x \) to four decimal places as instructed. The number 0.0245907 rounded to four decimal places is:\( x \approx 0.0246 \)

Key Concepts

Exponential Function

The exponential function is a fundamental mathematical function that grows rapidly. It is represented as \( e^x \), where \( e \) is a constant approximately equal to 2.71828. This function has unique properties, such as:

• Its rate of growth is proportional to its current value.
• It is continuous and differentiable, which means it's smooth and has a slope everywhere.
• It is one-to-one, meaning each input results in one unique output.

These characteristics make the exponential function crucial in various fields like biology for modeling population growth, and finance for calculating compound interest. In the given exercise, when facing the equation \( \ln x = -3.71 \), we apply the exponential function to both sides to find \( x \), changing it into \( x = e^{-3.71} \). This powerful tool transforms the problem from a logarithmic to an algebraic one, providing an effective solution path.

Inverse Functions

Inverse functions essentially reverse the effect of their corresponding functions. For any function \( f \), the inverse function, denoted as \( f^{-1} \), works by swapping the roles of the input and output. In mathematical terms, if \( y = f(x) \), then \( x = f^{-1}(y) \).The natural logarithm \( \ln(x) \) and the exponential function \( e^x \) are perfect examples of inverse functions:

• The natural logarithm finds which power the constant \( e \) must be raised to produce \( x \).
• The exponential function takes a logarithm result and returns the original base number.

In the context of solving \( \ln x = -3.71 \), recognizing that the natural logarithm is the inverse of the exponential function allows us to transition smoothly between the two. We utilize the relation \( x = e^y \) once we have \( y = \ln(x) \), effectively unwrapping the logarithm and solving for \( x \). This power of inverse functions is crucial any time we need to solve equations using transformations.

Scientific Calculator Usage

Handling mathematical calculations in equations like \( e^{-3.71} \), a scientific calculator becomes an invaluable tool. Here's how to use it efficiently:Start by ensuring that your calculator is in "scientific mode," which allows access to functions like exponents. To calculate \( e^{-3.71} \):

• Enter \(-3.71\) into the calculator.
• Press the \( e^x \) button to apply the exponential function to this input value.
• The resulting display provides the calculated value, around 0.0245907 in our case.
• Lastly, if needed, round this to four decimal places for precision, resulting in 0.0246.

Using a scientific calculator properly ensures accurate and rapid calculations, essential for both homework and real-world applications. Regular practice with these devices increases speed and confidence in tackling complex mathematical problems.