Problem 400 - Precalculus

Question

For the following exercises, use this scenario: A doctor prescribes 125 milligrams of a therapeutic diug 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 of the drug is approximately 2 hours.

Step 1: Understand the Problem

We are given that a drug decays by 30% each hour. We need to find the half-life, which is the time it takes for the drug amount to reduce to half its initial amount.

Step 2: Define the Decay Formula

The exponential decay formula is given by \( A = A_0 (1 - r)^t \), where \( A_0 \) is the initial amount, \( r \) is the decay rate, and \( t \) is the time in hours. Here, \( A_0 = 125 \)mg and \( r = 0.30 \).

Step 3: Set Up the Half-Life Equation

We need the final amount to be half of the initial amount. So, we set \( A = \frac{1}{2}A_0 \). Substituting, the equation becomes \( 0.5 \times 125 = 125 (0.7)^t \).

Step 4: Simplify the Equation

Divide both sides by 125 to isolate the decay term: \( 0.5 = (0.7)^t \).

Step 5: Solve for Time Using Logarithms

To solve for \( t \), take the natural logarithm of both sides: \( \ln(0.5) = \ln((0.7)^t) = t \cdot \ln(0.7) \).

Step 6: Calculate the Value of \( t \)

Solve for \( t \) by dividing both sides by \( \ln(0.7) \): \( t = \frac{\ln(0.5)}{\ln(0.7)} \). Calculate this to get \( t \approx 2.41 \) hours.

Step 7: Round To Nearest Hour

The calculated time is approximately 2.41 hours. Rounding to the nearest whole number, the half-life is 2 hours.

Key Concepts

Understanding Half-Life

Half-life is an important concept when dealing with exponential decay. It refers to the time required for a substance to reduce to half of its initial amount. For example, if a drug that initially weighs 125 milligrams has a half-life of 2 hours, after 2 hours, only 62.5 milligrams will remain.

Understanding the half-life helps in determining how frequently a drug needs to be administered to maintain its effectiveness. The concept is also widely used in other fields such as physics for radioactivity and biology in population studies.

• Half-life helps measure the 'speed' of a decay process.
• It is a constant for a given substance under specific conditions.
• Half-life does not mean the substance has been completely expelled in twice that time.

Knowing the half-life allows you to predict how long it will take for any initial amount to be reduced to a certain level, using the formula related to exponential decay.

Exploring Decay Rate

The decay rate is the rate at which a substance decreases in quantity over time. In the context of exponential decay, it is expressed as a percentage and is crucial for calculating how fast the substance decays.

In our scenario, the decay rate of the drug is 30%, meaning each hour, the drug's amount decreases by 30%. The decay rate is typically a decimal in formulas, such as 0.30 for 30%.

• A higher decay rate means the substance decays faster.
• It is important as it helps to understand how the substance's concentration decreases over time.
• The decay rate remains consistent unless the conditions change.

The decay process can be visualized with exponential graphs, illustrating how quickly the substance reduces, making calculations more intuitive.

The Role of Logarithms

Logarithms play a vital role in solving exponential decay problems, particularly for finding unknown variables such as time.

In our exercise, logarithms were used to solve for the time until the drug's amount reaches half of its initial quantity — this is where the half-life comes in. To find the half-life, we equate the logarithmic expressions derived from the exponential decay formula:

• Taking logarithms helps in linearizing the decay formula, making it easier to solve for the time.
• Using properties of logarithms, complex equations become simpler to solve.
• Natural logarithms ( ln ) are particularly helpful in problems involving exponential decay.

With logarithms, one can break down the mathematical complexities of decay problems and solve for variables that would otherwise be cumbersome to calculate manually.