Problem 22 - Algebra A Combined Function

Question

Practice using the exponential decay formula with half-lives by completing the table below. The first row has been completed for you. $$ \begin{array}{|c|c|c|c|c|c|} \hline \begin{array}{c} \text { Original } \\ \text { Amount } \end{array} & \begin{array}{c} \text { Half-Life } \\ \text { (in years) } \end{array} & \begin{array}{c} \text { Number } \\ \text { of Years } \end{array} & \begin{array}{c} \text { Time Intervals, } \boldsymbol{x}\left(\frac{\text { Years }}{\text { Half-Life }}\right) \\ \text { Rounded to Tenths if Needed } \end{array} & \begin{array}{c} \text { Final Amount after } \boldsymbol{x} \text { Time } \\ \text { Intervals (rounded to tenths) } \end{array} & \begin{array}{c} \text { Is Your Final Amount } \\ \text { Reasonable? } \end{array} \\ \hline 60 & 8 & 10 & \frac{10}{8}=1.25 & 25.2 & \text { yes } \\ \hline \text { a. } 200 & 12 & 36 & & & \\ \hline \text { b. } 200 & 12 & 40 & & & \\ \hline \end{array} $$

Step-by-Step Solution

Answer

Row a: Final Amount = 25, Reasonable: Yes; Row b: Final Amount ≈ 20.7, Reasonable: Yes.

Step 1: Calculate Time Intervals for Row a

To find the time intervals for row a, divide the number of years by half-life: \[\frac{36}{12} = 3\].

Step 2: Calculate Final Amount for Row a

Using the exponential decay formula, the final amount is calculated as: \[\text{Final Amount} = 200 \times \left(\frac{1}{2}\right)^3 = 200 \times 0.125 = 25\].

Step 3: Determine Reasonableness for Row a

As the original amount is 200 and after 3 half-lives it reduces to 25, the answer is reasonable.

Step 4: Calculate Time Intervals for Row b

To find the time intervals for row b, divide the number of years by half-life: \[\frac{40}{12} \approx 3.33\].

Step 5: Calculate Final Amount for Row b

Using the exponential decay formula, the final amount is calculated as: \[\text{Final Amount} = 200 \times \left(\frac{1}{2}\right)^{3.33} \approx 200 \times 0.1035 = 20.7\].

Step 6: Determine Reasonableness for Row b

As the original amount is 200, and we calculate approximately 20.7 after 3.33 half-lives, the answer is reasonable.

Key Concepts

Half-Life Calculations

Understanding half-life is crucial when dealing with exponential decay. The half-life of a substance is the time it takes for half of it to decay or reduce in quantity. Basically, it's a measure of how quickly a material or substance breaks down.
Half-life calculations are essential in fields like chemistry, biology, and physics, as they help in determining how long a substance remains active or potent.
For example, in the provided exercise, we look at two rows: row a with 36 years and row b with 40 years. By dividing the number of years by the half-life, we find the number of time intervals that have passed.

• For row a: \[\frac{36}{12} = 3\] time intervals.
• For row b:\[\frac{40}{12} \approx 3.33\] time intervals.

These calculations form the basis for further determining how much of a substance remains after these intervals.

Exponential Decay Formula

The exponential decay formula is a mathematical expression used to model the decline of a quantity over time. This formula is expressed as:
\[\text{Final Amount} = ext{Original Amount} \times \left(\frac{1}{2}\right)^n\]where

• \(n\) represents the number of half-life intervals.

In our exercise, we apply this formula to find out how much of the original quantity remains after a certain amount of time.
For row a, with 3 intervals, the final amount is:\[200 \times \left(\frac{1}{2}\right)^3 = 25\]For row b, with approximately 3.33 intervals, it results in:\[200 \times \left(\frac{1}{2}\right)^{3.33} \approx 20.7\]These calculations show how quickly the quantity decreases as time progresses.

Mathematical Reasoning

Mathematical reasoning is the logic we apply to solve problems systematically. When dealing with exponential decay, understanding the relationships between variables is key. This helps us reason through how a substance's amount changes over time.
In our scenario, we compare the original amounts with the calculated final amounts to determine whether our answers are reasonable. Checking if the final amount makes sense given the time intervals and half-life ensures our calculations are correct.
In row a, 200 reduces logically to 25 after 3 half-lives, and in row b, 20.7 after about 3.33 half-lives. Spotting these patterns boosts confidence in the results and confirms the application of the exponential decay formula. It’s about making connections and validating the data, showing that results are aligned with expectations.

Algebraic Problem Solving

Algebraic problem solving often involves breaking down the given problem into smaller, manageable steps. It requires a combination of understanding the mathematical principles at play and applying them effectively.
In our exercise, algebraic problem solving helps us:

• First, calculate time intervals using division.
• Second, apply the exponential decay formula to find how much of a substance remains.
• Finally, verify whether the results are reasonable by comparing it to previous knowledge or expected outcomes.

This structured approach, using calculations and logical deductions, makes algebra an indispensable tool in problem solving. By methodically applying algebra, such as in exponential decay exercises, we can solve complex real-world problems with relative ease.