Problem 72 - College Algebra with Corequisite Support

Question

For the following exercises, solve each equation by rewriting the exponential expression using the indicated logarithm. Then use a calculator to approximate the variable to 3 decimal places. \(50 e^{-0.12 t}=10\) using the natural log

Step-by-Step Solution

Answer

The solution is approximately \( t = 13.412 \).

Step 1: Isolate the Exponential Term

Start by isolating the exponential term on one side of the equation. We have the equation \( 50 e^{-0.12 t} = 10 \). To isolate \( e^{-0.12 t} \), divide both sides by 50: \( e^{-0.12 t} = \frac{10}{50} \). Simplifying the right side gives us \( e^{-0.12 t} = 0.2 \).

Step 2: Apply Natural Logarithm

Take the natural logarithm (ln) on both sides of the equation to remove the base \( e \). This results in \( \ln(e^{-0.12 t}) = \ln(0.2) \). Since \( \ln(e^x) = x \), this simplifies to \( -0.12 t = \ln(0.2) \).

Step 3: Solve for the Variable t

To find \( t \), divide both sides of the equation by \(-0.12\): \( t = \frac{\ln(0.2)}{-0.12} \). Calculate \( \ln(0.2) \) using a calculator and solve for \( t \).

Step 4: Approximate Using a Calculator

Calculate \( \ln(0.2) \) which is approximately \(-1.6094379\). Then, calculate \( t = \frac{-1.6094379}{-0.12} \) which evaluates to \( t \approx 13.412 \).

Key Concepts

Natural Logarithms

Natural logarithms are logarithms with the base of Euler's number, which is approximately 2.71828, commonly denoted as \( e \). They are represented as \( \ln(x) \), which is used frequently in mathematics to solve equations involving exponential growth or decay.
In the context of our problem, we apply the natural logarithm to both sides of the equation \( e^{-0.12 t} = 0.2 \). Using the property \( \ln(e^x) = x \), we simplify \( \ln(e^{-0.12 t}) \) to just \( -0.12 t \).

• One main use of natural logarithms is to "undo" the exponential function so we can solve for a variable contained within the exponent.
• This helps us isolate variables efficiently and forms the basis of logarithmic functions.

Understanding and applying natural logarithms is essential not just for solving math problems but for various fields like science and engineering, where modeling exponential growth or decay is prevalent.

Isolation of Variables

In solving equations, one effective strategy is the isolation of variables. This involves rearranging the equation so that the variable you want to solve for is on one side of the equation, by itself.
In the given problem, we begin with \( 50 e^{-0.12 t} = 10 \). To isolate the term involving the variable, \( e^{-0.12 t} \), we divide both sides by 50. This reduces the equation to \( e^{-0.12 t} = 0.2 \).

• The goal is to have the variable \( t \) by itself, which means we need to eliminate other numbers and exponents around it.
• By systematically applying mathematical operations like division and taking logarithms, isolation becomes straightforward.

Once isolated, solving the rest of the equation becomes more manageable. This step is foundational because it simplifies complex equations into more solvable forms.

Solving Equations Step-by-Step

Solving equations step-by-step ensures clarity and precision. We follow a sequence of logical steps to arrive at the solution.
In our case, after isolating the variable term, we take the natural logarithm to further simplify the equation. As a result, \( -0.12 t = \ln(0.2) \) is achieved, allowing us to solve for \( t \) straightforwardly.

• This method is crucial as it helps break down potentially intimidating problems into smaller, more manageable parts.
• Each step builds upon the last, ensuring that no errors are made, and any mistakes can be easily identified and corrected.

This structured approach is not only helpful for solving exponential equations but is a powerful tool for tackling various mathematical challenges.

Using Calculators for Approximation

When working with complex numbers, especially in logarithmic or exponential equations, approximations using calculators are often necessary.
For example, calculating \( \ln(0.2) \) results in roughly \(-1.6094379\). Using this approximation, we solve for \( t \) by computing \( \frac{-1.6094379}{-0.12} \), resulting in approximately \( 13.412 \).

• Calculators provide quick and accurate approximations for logarithms and other functions that are cumbersome to compute by hand.
• Approximations are typically presented to a set number of decimal places for clarity and consistency, as in this problem where we used 3 decimal places.

While calculators are helpful, it is always essential to understand the underlying mathematics of what you're calculating. This ensures not just accuracy but also insight into why certain results make sense or may need reevaluation.