Problem 33 - College Algebra

Question

Use this scenario: A tumor is injected with 0.5 grams of Iodine-125, which has a decay rate of 1.15% per day. A scientist begins with 250 grams of a radioactive substance. After 250 minutes, the sample has decayed to 32 grams. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest minute, what is the half-life of this substance?

Step-by-Step Solution

Answer

The half-life is approximately 84 minutes.

Step 1: Understand the Decay Formula

The exponential decay of a substance can be represented by the formula \( A = A_0 e^{-kt} \), where \( A_0 \) is the initial amount of substance, \( A \) is the amount remaining after time \( t \), and \( k \) is the decay constant. Our goal is to find the decay constant \( k \).

Step 2: Write the Given Information

From the problem, we have an initial amount \( A_0 = 250 \) grams and an amount \( A = 32 \) grams remaining after \( t = 250 \) minutes.

Step 3: Solve for the Decay Constant \( k \)

Substitute the known values into the decay formula: \[ 32 = 250 e^{-250k} \]Now, solve for \( k \): \[ e^{-250k} = \frac{32}{250} \] Take the natural logarithm of both sides:\[ -250k = \ln\left(\frac{32}{250}\right) \] \[ k = -\frac{\ln\left(\frac{32}{250}\right)}{250} \] Evaluate the expression to find \( k \).

Step 4: Calculate \( k \)

Firstly, calculate \( \frac{32}{250} = 0.128 \). Then, \( \ln(0.128) \approx -2.0541237 \). Thus, \( k = -\frac{-2.0541237}{250} \approx 0.0082165 \) (rounded to five significant digits).

Step 5: Write the Exponential Equation

Now that \( k \) is known, write the exponential decay equation: \[ A = 250 e^{-0.0082165t} \]

Step 6: Determine the Half-Life

The half-life is the time \( t \) when \( A = \frac{A_0}{2} \). Set up the equation for half-life: \[ 125 = 250 e^{-0.0082165t} \] Solve for \( t \):\[ e^{-0.0082165t} = 0.5 \] Take the natural logarithm of both sides:\[ -0.0082165t = \ln(0.5) \] \[ t = \frac{\ln(0.5)}{-0.0082165} \] Calculate \( t \).

Step 7: Calculate the Half-Life

Compute \( \ln(0.5) \approx -0.693147 \). Thus, \( t = \frac{-0.693147}{-0.0082165} \approx 84.4 \). Rounding to the nearest minute, the half-life is approximately 84 minutes.

Key Concepts

Decay Constant

The decay constant, denoted as \( k \), is a crucial part of the exponential decay formula. It tells us how quickly a substance is decaying over time. In our problem, the decay constant is used to determine how fast the radioactive substance is losing its mass.
To find \( k \), we start with the decay formula:

• \( A = A_0 e^{-kt} \)
• Where \( A_0 \) is the initial mass, \( A \) is the mass after time \( t \), and \( k \) is our decay constant.

By plugging in the values provided:

• Initial mass \( A_0 = 250 \) grams
• Remaining mass \( A = 32 \) grams after \( t = 250 \) minutes

We rearrange to solve for \( k \), first isolating \( k \) by dividing and then taking the natural logarithm:

• \( -250k = \ln\left(\frac{32}{250}\right) \)
• After calculation, we find \( k \approx 0.0082165 \)

This shows the rate at which the substance decays every minute.

Half-Life

Half-life is the time it takes for half of the radioactive substance to decay. It is a useful measure to understand the stability and longevity of a substance over time. In our exercise, calculating the half-life is important for predicting how quickly the substance loses half of its mass.
We start with the exponential decay model:

• \( A = A_0 e^{-kt} \)

To find the half-life, set \( A \) to half of \( A_0 \):

• \( A = \frac{A_0}{2} \)

Thus the equation becomes:

• \( 125 = 250 e^{-0.0082165t} \)

Solving for \( t \) involves taking the natural logarithm:

• \( -0.0082165t = \ln(0.5) \)
• Simplify to find \( t \approx 84.4 \)
• Rounding gives \( t \approx 84 \) minutes

This means every 84 minutes, half of the substance remains.

Exponential Equation

An exponential equation is a mathematical expression where a variable appears in the exponent. In the context of our problem, the exponential equation models the decay of a radioactive substance over time.
The general form is:

• \( A = A_0 e^{-kt} \)

Where:

• \( A \) is the mass at time \( t \)
• \( A_0 \) is the initial mass
• \( k \) is the decay constant

For our specific example:

• The exponential decay equation becomes \( A = 250 e^{-0.0082165t} \)
• This models the decay from 250 grams to 32 grams over 250 minutes

The equation helps us predict the remaining amount of substance at any given time \( t \).

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a mathematical function used to solve problems involving exponential decay. It is the inverse operation of exponentiation.
When determining the decay constant \( k \) or the half-life \( t \), the natural logarithm helps to simplify equations by "undoing" the exponential function.

• For example, in the decay calculation, \( e^{-250k} = \frac{32}{250} \), taking the natural logarithm of both sides gives \( -250k = \ln\left(\frac{32}{250}\right) \)
• For the half-life, take \( \ln \) on both sides to solve \( e^{-0.0082165t} = 0.5 \)

The natural logarithm is essential for these calculations, providing a straightforward way to work through exponential equations.