Problem 28 - College Algebra

Question

For the following exercises, use this scenario: A doctor prescribes 125 milligrams of a therapeutic drug that decays by about \(30 \%\) each hour. To the nearest hour, what is the half-life of the drug?

Step-by-Step Solution

Answer

The half-life is approximately 2 hours.

Step 1: Understanding the Problem

We need to determine the time it takes for the drug's amount to reduce to half of its original quantity. The drug initially is 125 mg and any transition should be worked with this initial value and the decay rate.

Step 2: Setting Up the Formula

The problem states that the drug decays at a rate of 30% per hour. We can use the exponential decay formula: \[ y(t) = y_0 imes (1 - r)^t \] where \( y(t) \) is the remaining amount at time \( t \), \( y_0 \) is the initial amount of the drug, and \( r \) is the decay rate.

Step 3: Substituting Known Values

Here \( y_0 = 125 \) mg and \( r = 0.3 \). We need to find \( t \) such that the amount becomes half of 125 mg, i.e., 62.5 mg. The equation becomes \[ 62.5 = 125 imes (1 - 0.3)^t \]

Step 4: Simplifying the Equation

Divide both sides by 125: \[ \frac{62.5}{125} = (0.7)^t \] This simplifies to \[ 0.5 = (0.7)^t \]

Step 5: Using Logarithms to Solve for Time

Apply logarithms to both sides to solve for \( t \): \[ \log(0.5) = t \log(0.7) \] Then rearrange to find \( t \): \[ t = \frac{\log(0.5)}{\log(0.7)} \]

Step 6: Calculating the Half-Life

Using a calculator, find the values:\[ \log(0.5) \approx -0.3010 \] and \( \log(0.7) \approx -0.1549 \). Therefore, \[ t \approx \frac{-0.3010}{-0.1549} \approx 1.9440 \]

Key Concepts

Half-Life Calculation

Calculating the half-life of a substance is a key application of exponential decay. In scenarios involving substances like drugs, which reduce over time, understanding half-life is crucial.
In our example, we start with 125 mg of a drug, and we want to know when it will be halved in quantity, down to 62.5 mg.
The half-life is the time it takes for any material to decrease to half of its starting amount. Here, using our exponential decay formula with the given rate of decay, we set the equation such that the remaining quantity is half of what we began with. This allows us to solve for the specific time period.

• Identify the initial amount and rate.
• Set the formula with half the initial amount as the target.
• Solve for time using the decay information.

Accurate calculations of half-life help in predicting dosages and scheduling the next administration in medical scenarios.

Exponential Functions

Exponential functions are mathematical expressions where the variable appears in the exponent. This makes them versatile in representing growth and decay.
In the context of our drug decay problem, the function models the drug's decreasing quantity over time.
The general form of an exponential decay function is:

• Formula: \( y(t) = y_0 \times (1 - r)^t \)
• Where:
  • \( y(t) \) is the amount remaining at time \( t \)
  • \( y_0 \) is the initial amount
  • \( r \) is the decay rate

Here, we see how exponential functions describe the drug's reduction from 125 mg at time zero.
Exponential functions provide a precise model for a variety of phenomena, making them critical in both academic studies and real-world applications.

Logarithms

Logarithms are the inverse operations of exponentiation and are essential in solving equations where the exponent is unknown.
In our problem, once the equation for decay is set, logarithms enable us to deduce the time variable.
For instance, solving \( 0.5 = (0.7)^t \) requires understanding that you can use logarithms to "take down" the exponent so that it can be isolated:

• Applying log to both sides: \( \log(0.5) = t \log(0.7) \)
• Rearranging gives: \( t = \frac{\log(0.5)}{\log(0.7)} \)

This manipulation allows for numerical calculation of \( t \), using known logarithmic values.
By using logarithms, we can manage the complexity of exponential equations, unlocking solutions where the unknown is "trapped" in the exponent.