Problem 582 - Precalculus

Question

A radiation safery officer is working with 112 grams of a radioactive substance. After 17 days, the sample has decayed to 80 grams. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest day, what is the half-life of this substance?

Step-by-Step Solution

Answer

The exponential equation is \( A(t) = 112 \cdot e^{-0.02071t} \), and the half-life is approximately 33 days.

Step 1: Understand the Exponential Decay Model

The exponential decay of a substance is modeled by the equation \( A(t) = A_0 \cdot e^{kt} \), where \( A(t) \) is the amount remaining after time \( t \), \( A_0 \) is the initial amount, \( k \) is the decay constant, and \( t \) is time.

Step 2: Set Up Given Values

We know \( A_0 = 112 \) grams (initial amount) and \( A(17) = 80 \) grams (amount after 17 days). We need to solve for the decay constant \( k \).

Step 3: Substitute Values into the Decay Equation

Substitute the known values into the equation: \( 80 = 112 \cdot e^{17k} \). We will solve this equation for \( k \).

Step 4: Solve for the Decay Constant \( k \)

Divide both sides by 112: \( \frac{80}{112} = e^{17k} \). Taking the natural logarithm of both sides: \( \ln\left(\frac{80}{112}\right) = 17k \). Solve for \( k \): \( k = \frac{\ln\left(\frac{80}{112}\right)}{17} \).

Step 5: Calculate the Decay Constant

Perform the calculations: \( k \approx \frac{\ln(0.71429)}{17} \approx -0.02071 \). This gives us the decay constant \( k \approx -0.02071 \).

Step 6: Write the Exponential Equation

With \( A_0 = 112 \) and \( k \approx -0.02071 \), the exponential decay equation is \( A(t) = 112 \cdot e^{-0.02071t} \).

Step 7: Determine Half-Life

The half-life is the time \( t \) when the substance decays to half its initial amount: \( \frac{112}{2} = 56 \). Set \( 56 = 112 \cdot e^{-0.02071t} \) and solve for \( t \).

Step 8: Solve for the Half-Life

Divide both sides by 112: \( 0.5 = e^{-0.02071t} \). Take the natural logarithm of both sides: \( \ln(0.5) = -0.02071t \). Solve for \( t \): \( t = \frac{\ln(0.5)}{-0.02071} \).

Step 9: Calculate and Round Half-Life

Perform the calculation: \( t \approx \frac{-0.69315}{-0.02071} \approx 33.47 \). Round to the nearest day: the half-life is approximately 33 days.

Key Concepts

Half-Life Calculation

The half-life of a substance is the time it takes for half of it to decay. In radioactive decay, it's a crucial concept to understand how quickly a material diminishes. To find the half-life, we use the exponential decay equation.
First, set the time it takes for the material to be reduced to half its original amount. In the equation for decay, this means when the initial amount is reduced by half, i.e., from 112 grams to 56 grams in our case.
The half-life equation can thus be set using the equation: \[ \frac{A_0}{2} = A_0 \cdot e^{kt} \] So, after cancellation, it simplifies to:\[ e^{kt} = \frac{1}{2} \]From this equation, if you know the decay constant \( k \), solving for \( t \) will give you the half-life. Remember, the use of natural logarithms is crucial to isolate \( t \). By substituting \( k = -0.02071 \) (our decay constant) into the equation and solving, we find the half-life to be approximately 33 days after rounding.

Decay Constant

The decay constant, denoted as \( k \), is a vital part of understanding exponential decay processes. It indicates the rate at which a substance decays over time.
To calculate \( k \), one must analyze the time taken for a substance to reduce from its initial mass to a given lesser mass.
For the calculation: use the logarithmic form of the decay equation. In our example, after substituting 80 for the remaining amount after 17 days, the setup is:\[ 80 = 112 \cdot e^{17k} \] Divide both sides by 112:\[ \frac{80}{112} = e^{17k} \] Now, to isolate \( k \), take the natural logarithm of both sides:\[ \ln\left( \frac{80}{112} \right) = 17k \] This transformation allows us to solve for \( k \), giving us the value \( k \approx -0.02071 \). This signifies a mild decay rate, which affects how the material halves over given time.

Natural Logarithm

Natural logarithms, represented by \( \ln \), play a key role in solving exponential decay problems. They help convert exponential expressions into polynomial forms for easier manipulation.
Taking the natural logarithm helps simplify equations because it is the inverse of the exponential function \( e^x \). If \( y = e^x \), then taking the natural log, \( \ln \), gets us back to \( x \).In half-life calculations and finding the decay constant, \( \ln \) changes the exponential decay form into a manageable equation. For instance, turning \( e^{17k} = \frac{80}{112} \) into \( \ln\left( \frac{80}{112} \right) = 17k \) drastically simplifies the solving process.
This technique not only aids in finding decay constants but also effectively determines how long it will take for substances like radioactive materials to diminish by half. Assessing natural logs is fundamental in stripping away complexities inherent in understanding exponential decay.