Category Archives: 6.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models

5.5 Applications of Exponential and Logarithmic Functions, 6.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models

Application: Exponential Growth and Decay (Half-life)

Example: The half life of radium is 1690 years. If 50 grams are present now, how much will be present in 630 years?

Solution:  There is a two part process to this problem.  Part 1: Use some of the information to find the decay rate of radium.  Part 2: Answer the question using the rest of the given information.

Part 1:  Find the decay rate of radium.

Since we are using an exponential model for this problem we should be clear on the parts of the exponential decay model.

Exponential Decay Model

A=A _0 e^{kt}

A _0 is the initial amount

k is the decay rate

t is the time

A is the amount after t time has passed

Since radium has a half life of 1690 years, we know that after 1690 years there will be half of the initial amount of radium left.  This allows me to establish a relationship between the initial amount and the amount after 1690.   The amount after 1690 year is half of the initial amount.   A=1 /2 A _0 when the t=1690.

Substitute these values into the exponential decay formula and solve for k.

 

Exponential Decay Formula
A=A _0 e^{kt}
Make a substitution for A and t since it is known that the half-life is 1690 years A=1 /2 A _0 and t=1690
A=A _0 e^{kt}
1 /2 A _0=A _0 e^{k(1690)}
Solve for the decay rate k:
Start by dividing both sides by the coefficient to isolate the exponential factor
1 /2 A _0=A _0 e^{k(1690)}
{1 /2 A _0}/A_0={A _0 e^{k(1690)}}/A_0
1 /2= e^{k(1690)}
Solve for the decay rate k:
Take the natural log of both sides to get k out of the exponent
1 /2= e^{k(1690)}
ln(1 /2)= ln(e^{k(1690)})
Solve for the decay rate k:
Use the power rule for logarithms to get k out of the exponent
ln(1 /2)= ln(e^{k(1690)})
ln(1 /2)= k(1690)ln(e)
Solve for the decay rate k:
Simplify ln e = 1
ln(1 /2)= k(1690)ln(e)
ln(1 /2)= k(1690)(1)
ln(1 /2)= k(1690)
Solve for the decay rate k:
Solve for k by dividing by 1690 on both sides
ln(1 /2)= k(1690)
{ln(1 /2)}/1690= {k(1690)}/1690
{ln(1 /2)}/1690= k
-0.000410 approx k

Using only the information about radium having a half-life of 1690 years I have found the decay rate for radium.

k={ln(1 /2)}/1690
k approx -0.000410

Note: Although I have put an approximation for k here, try not to round until the very last step.

Part 2:  Answer the question using the rest of the given information.

Given information: If 50 grams are present now, how much will be present in 630 years?

With this information I can identify the initial amount of radium as 50 grams and the time to be 630 years.  Symbolically that is A _0=50 when the t=630.  Substitute the given information and the decay rate k found in part 1 to the exponential decay formula.

Exponential Decay Formula
A=A _0 e^{kt}
Make a substitution for initial amount (A_0), time (t), and the decay rate (k).
A _0=50 , t=630, and k={ln(1 /2)}/1690
A=A _0 e^{kt}
A=50 e^{630{ln(1 /2)}/1690}
Type the expression into your calculator and round to the thousandth place.
A=50 e^{630{ln(1 /2)}/1690}
A approx 38.615

A approx 38.615

38.615 grams will be present 630 years later is 50 grams are present initially.

Here is a video example that is similar to the above example and it shows how to enter information in the calculator.