Problem 39 - Intermediate Algebra

Question

The number of grams of a certain radioactive substance present after \(t\) hours is given by the equation \(Q=\) \(Q_{0} e^{-0.45 t}\), where \(Q_{0}\) represents the initial number of grams. How long will it take 2500 grams to be reduced to 1250 grams?

Step-by-Step Solution

Answer

It will take about 1.54 hours for 2500 grams to be reduced to 1250 grams.

Step 1: Write Down the Given Information

The initial quantity of the substance is \( Q_0 = 2500 \) grams and the final quantity is 1250 grams. We need to find the time \( t \) it takes for the substance to reduce to 1250 grams.

Step 2: Set Up the Equation

Use the equation \( Q = Q_0 e^{-0.45 t} \). Substitute \( Q = 1250 \) and \( Q_0 = 2500 \) to get the equation: \[ 1250 = 2500 e^{-0.45t} \]

Step 3: Isolate the Exponential Term

Divide both sides of the equation by 2500 to isolate the exponential term:\[ \frac{1250}{2500} = e^{-0.45t} \]This simplifies to: \[ 0.5 = e^{-0.45t} \]

Step 4: Solve for Time Using Natural Logarithms

Take the natural logarithm (ln) of both sides to solve for \( t \):\[ \ln(0.5) = \ln(e^{-0.45t}) \]Using the property \( \ln(e^x) = x \), we get:\[ \ln(0.5) = -0.45t \]

Step 5: Calculate Time

Solve for \( t \) by dividing both sides by -0.45:\[ t = \frac{\ln(0.5)}{-0.45} \]Calculate the result using a calculator.

Key Concepts

Exponential Functions

Exponential functions are essential in modeling situations where quantities grow or decay at a constant relative rate over time, such as in radioactive decay. In mathematics, an exponential function is expressed in the form \( f(x) = a \, e^{bx} \), where:

• \( a \) is a constant, which stands for the initial value or amount.
• \( e \) represents the base of the natural logarithm (approximately 2.718).
• \( b \) is the growth (or decay) rate constant.
• \( x \) is the independent variable, often representing time.

Radioactive decay is a common context for exponential functions because radioactive substances decrease at a rate proportional to their current amount. This means as time progresses, the rate of decay slows down because less of the substance is present to decay. By understanding the general form of exponential functions, you can model any situation with constant proportional growth or decay.
In our problem, the decay equation is \( Q = Q_0 \, e^{-0.45t} \), emphasizing how the quantity \( Q \) changes over time \( t \) based on the initial amount \( Q_0 \) and a rate of decay given by \(-0.45\).

Natural Logarithms

Natural logarithms, denoted as \( \ln \), are inverse operations to exponential functions. They help us solve equations where the variable is in the exponent. The base of the natural logarithm is \( e \), and the natural log of a number \( x \) is written as \( \ln(x) \).
In situations involving exponential decay, like our exercise, natural logarithms unravel the exponent to solve for an unknown variable, such as time. Once we isolate the exponential term, we can apply the natural logarithm to both sides of an equation. This leverages the property \( \ln(e^x) = x \), effectively canceling the exponential and bringing down the exponent:

• Given, \( 0.5 = e^{-0.45t} \), applying the natural logarithm yields: \( \ln(0.5) = \ln(e^{-0.45t}) \), which simplifies to \( \ln(0.5) = -0.45t \).

This simplification allows for solving \( t \). Using natural logarithms is key in manipulating exponential equations for easy handling of decay and growth processes involving radioactive substances.

Half-Life Calculations

The half-life of a substance is the time required for half of a given amount of radioactive substance to decay. Calculating half-life is crucial in understanding the decay process over time. It is distinct from finding when the substance reaches half its initial quantity due to its relation with exponential decay.
To calculate half-life, we use the continuous decay formula: \( Q = Q_0 \, e^{-kt} \). If we set \( Q = \frac{Q_0}{2} \) by definition of half-life, the equation becomes:

• \( \frac{Q_0}{2} = Q_0 \, e^{-kt} \)
• Dividing both sides by \( Q_0 \) simplifies it to \( \frac{1}{2} = e^{-kt} \)
• Taking the natural logarithm: \( \ln(\frac{1}{2}) = -kt \)
• Solve for \( t \) (half-life \( T_{1/2} \)): \( T_{1/2} = \frac{\ln(0.5)}{-k} \)

This formula shows that the half-life depends solely on the decay constant \( k \) and is independent of the initial quantity \( Q_0 \). Understanding half-life helps in fields like nuclear physics and environmental science, where it's crucial to know how long a radioactive substance persists and alters over time.