Problem 11 - Applied Calculus

Question

Solve for \(t\) using natural logarithms. $$40=100 e^{-0.03 t}$$

Step-by-Step Solution

Answer

The solution is approximately \( t \approx 30.54 \).

Step 1: Isolate the Exponential Term

Begin by isolating the exponential term in the equation. Move 100 to the other side by dividing both sides by 100. This results in: \[ \frac{40}{100} = e^{-0.03t} \] Simplifying the left side gives: \[ 0.4 = e^{-0.03t} \]

Step 2: Apply the Natural Logarithm

To solve for \( t \), use the natural logarithm on both sides of the equation to eliminate the exponential function: \[ \ln(0.4) = \ln(e^{-0.03t}) \] Using the property of logarithms, \( \ln(e^x) = x \), the equation simplifies to: \[ \ln(0.4) = -0.03t \]

Step 3: Solve for \( t \)

Now, isolate \( t \) by dividing both sides by \(-0.03\): \[ t = \frac{\ln(0.4)}{-0.03} \] Calculate the value of \( t \) using a calculator: \[ t \approx \frac{-0.9163}{-0.03} \] \[ t \approx 30.54 \]

Key Concepts

Understanding Exponential Equations

Exponential equations are a crucial concept in mathematics, commonly appearing in scenarios involving growth and decay. They are equations in which variables appear as exponents. For example, in the equation given in our exercise, the term \( e^{-0.03t} \) is an exponential expression. Here, \( e \) is the base of the natural logarithm, approximately equal to 2.718. The variable \( t \) appears in the exponent. These types of equations are used to model a variety of real-world situations, including population growth, radioactive decay, and interest calculations. To solve exponential equations, the aim is usually to isolate the term with the variable in the exponent, which often involves using logarithms to "bring down" the exponent to the variable level. Once this is done, ordinary algebra can take over to solve for the variable. This step is significant, as it transforms an exponential problem into a more manageable linear one.

Key Logarithmic Properties

Logarithms are the inverse operations of exponentiation, a critical tool in solving exponential equations. They serve an essential function in algebra, allowing us to manipulate expressions where the unknowns are in the exponents.A few key properties of logarithms make them particularly useful:

• \( \log_b(b^x) = x \): This property states that the logarithm of a number where the base matches the exponent’s base simplifies directly to the exponent.
• \( \ln(e^x) = x \): Logarithms with the natural base, \( e \), simplify similarly.
• The product, quotient, and power rules help simplify more complex logarithmic expressions.

In our exercise, we used the natural logarithm, \( \ln \), because the base of the exponential was \( e \). Applying \( \ln \) simplified our equation by removing the exponential and isolating the variable \( t \) as shown: \( \ln(0.4) = -0.03t \). Understanding these properties is vital for effectively solving mathematical problems involving exponential equations.

Enhancing Mathematical Problem-Solving Skills

Mathematical problem-solving is a foundational skill that extends beyond math itself. It involves logic, patience, and a step-by-step approach to finding solutions.When faced with equations, especially those involving logarithms and exponentials, keep the following strategies in mind:

• **Break the problem into smaller parts:** As shown in the original solution, start by isolating the terms to simplify the equation.
• **Choose the right tools:** Know which mathematical properties and functions, such as natural logarithms, are most effective.
• **Check your work:** Always go back and substitute your solution into the original problem to verify its correctness.

These strategies help in approaching problems methodically, allowing you to solve similar challenges more confidently and accurately. In the given exercise, each step was crucial—from isolating the exponential term to applying the natural logarithm—to reach the final solution of \( t \approx 30.54 \). By honing these skills, you will find yourself solving complex problems more efficiently.