Suppose the amount of a radioactive substance is cut in half every 25 years. If there was originally 500 grams of the substance, could you write a function representing the amount of the substance after \begin{align*}x\end{align*} years? How much of the substance would there be after 100 years? Will the amount of the substance ever reach 0 grams?

### Exponential Decay

In the last Concept, we learned how to solve expressions that modeled exponential growth. Now, we will be learning about exponential decay functions.

Recall that the general form of an exponential function is \begin{align*}y=a (b)^x\end{align*}, where \begin{align*}a=\end{align*} initial value and\begin{align*}b = growth \ factor\end{align*}

In exponential decay situations, the growth factor must be a fraction between zero and one.

\begin{align*}0 < b < 1\end{align*}

#### Let's use an exponential function to answer the following problem:

For her fifth birthday, Nadia’s grandmother gave her a full bag of candy. Nadia counted her candy and found out that there were 160 pieces in the bag. Nadia loves candy, so she ate half the bag on the first day. Her mother told her that if she continues to eat at that rate, it will be gone the next day and she will not have any more until her next birthday. Nadia devised a clever plan. She will always eat half of the candy that is left in the bag each day. She thinks that she will get candy every day and her candy will never run out. How much candy does Nadia have at the end of the week? Would the candy really last forever?

Make a table of values for this problem.

Day |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
---|---|---|---|---|---|---|---|---|

# of Candies |
160 | 80 | 40 | 20 | 10 | 5 | 2.5 | 1.25 |

You can see that if Nadia eats half the candies each day, then by the end of the week she has only 1.25 candies left in her bag.

Write an equation for this exponential function. Nadia began with 160 pieces of candy. In order to get the amount of candy left at the end of each day, we keep multiplying by \begin{align*}\frac{1}{2}\end{align*}. Because it is an exponential function, the equation is:

\begin{align*}y=160 \cdot \frac{1}{2}^x\end{align*}

#### Now, let's graph the following exponential functions:

- \begin{align*}y=5 \cdot \left(\frac{1}{2}\right)^x\end{align*}

Start by making a table of values. Remember when you have a number to the negative power, you are simply taking the reciprocal of that number and taking it to the positive power. Example: \begin{align*}\left(\frac{1}{2}\right)^{-2} = \left(\frac{2}{1}\right)^2 = 2^2\end{align*}.

\begin{align*}x\end{align*} | \begin{align*}y=5 \cdot \left(\frac{1}{2}\right)^x\end{align*} |
---|---|

–3 | \begin{align*}y =5 \left(\frac{1}{2}\right)^{-3}=40\end{align*} |

–2 | \begin{align*}y =5 \left(\frac{1}{2}\right)^{-2}=20\end{align*} |

–1 | \begin{align*}y =5 \left(\frac{1}{2}\right)^{-1}=10\end{align*} |

0 | \begin{align*}y =5 \left(\frac{1}{2}\right)^0=5\end{align*} |

1 | \begin{align*}y =5 \left(\frac{1}{2}\right)^1=\frac{5}{2}\end{align*} |

2 | \begin{align*}y =5 \left(\frac{1}{2}\right)^2=\frac{5}{4}\end{align*} |

Now graph the function.

Using the Property of Negative Exponents, the equation can also be written as \begin{align*}5 \cdot 2^{-x}\end{align*}.

- The functions \begin{align*}y=4^x\end{align*} and \begin{align*}y=4^{-x}\end{align*} on the same coordinate axes.

Here is the table of values and the graph of the two functions.

Looking at the values in the table, we see that the two functions are “reverse images” of each other in the sense that the values for the two functions are reciprocals.

\begin{align*}x\end{align*} | \begin{align*}y=4^x\end{align*} | \begin{align*}y=4^{-x}\end{align*} |
---|---|---|

–3 | \begin{align*}y=4^{-3} = \frac{1}{64}\end{align*} | \begin{align*}y=4^{-(-3)} = 64\end{align*} |

–2 | \begin{align*}y=4^{-2} = \frac{1}{16}\end{align*} | \begin{align*}y=4^{-(-2)} = 16\end{align*} |

–1 | \begin{align*}y=4^{-1} = \frac{1}{4}\end{align*} | \begin{align*}y=4^{-(-1)} = 4\end{align*} |

0 | \begin{align*}y=4^0 = 1\end{align*} | \begin{align*}y=4^{-(0)} = 1\end{align*} |

1 | \begin{align*}y=4^1 = 4\end{align*} | \begin{align*}y=4^{-(1)} =\frac{1}{4}\end{align*} |

2 | \begin{align*}y=4^2 = 16\end{align*} | \begin{align*}y=4^{-(2)} =\frac{1}{16}\end{align*} |

3 | \begin{align*}y=4^3 = 64\end{align*} | \begin{align*}y=4^{-(3)} = \frac{1}{64}\end{align*} |

Here is the graph of the two functions. Notice that these two functions are mirror images if the mirror is placed vertically on the \begin{align*}y-\end{align*}axis.

### Examples

#### Example 1

Earlier, you were told that the amount of a radioactive substance is cut in half every 25 years. If there was originally 500 grams of the substance, what is the function representing the amount of the substance after \begin{align*}x\end{align*} years? How much of the substance would there be after 100 years? Will the amount of the substance ever reach 0 grams?

The initial amount in this situation is 500 grams. If the amount of the substance is cut in half every 25 years then the growth factor is \begin{align*}\frac{1}{2}\end{align*}. You would assume that the function to represent this situation would be:

\begin{align*}y = 500 \cdot (\frac{1}{2})^x\end{align*}

with \begin{align*}x\end{align*} representing the number of years and \begin{align*}y\end{align*} representing the amount of the substance yet.

However, based on the situation, you know that after 25 years, there should be 250 grams of the substance left. If you plug 25 in for \begin{align*}x\end{align*}:

\begin{align*}y &= 500 \cdot (\frac{1}{2}^25\\ &=500 \cdot \frac{1}{33554432}\\ &=\frac{500}{33554432}\end{align*}

This is not the answer that we should get. The issue is in the exponent. Because the amount of substance is cut in half every 25 years, the exponent should increase in intervals of 25. Therefore the exponent should be \begin{align*}\frac{1}{25}x\end{align*}.

The equation that represents this situation is:

\begin{align*}y = 500 \cdot (\frac{1}{2})^{\frac{1}{25}x}\end{align*}

To find the amount of the substance that will be left after 100 years, plug in 100 to the equation.

\begin{align*}y &= 500 \cdot (\frac{1}{2})^{\frac{1}{25}(100)}\\
&=500 \cdot (\frac{1}{2})^{4}\\
&=500 \cdot \frac{1}{16}\\
&= 31.25\end{align*}

There will be 31.25 grams left after 100 years.

The amount will never reach zero because it will keep on getting cut in half but there is no number that can be divided 2 to get 0 apart from 0. There are an infinite amount of decimals and the number will continue to get smaller and smaller but there will never be 0 grams of the substance.

#### Example 2

If a person takes 125 milligrams of a drug, and after the full dose is absorbed into the bloodstream, there is only 70% left after every hour, write a function that gives the concentration left in the bloodstream after \begin{align*}t\end{align*} hours. What is the concentration of the drug in the bloodstream after 3 hours?

This will be a decay function in the form \begin{align*}f(t)=a\cdot b^t\end{align*}. We know the initial value is 125 milligrams. After one hour we multiply that by 0.70 to find 70% of 125. After the second hour, we multiply by 0.7 again and so on:

\begin{align*} \text{The initial dose.} && 125&=125 \\ \text{After one hour.} && 125 \cdot 0.7&=87.5\\ \text{After two hours.} && 125 \cdot 0.7 \cdot 0.7=125 \cdot 0.7^2&=61.25\\ \text{After three hours.} && 125 \cdot 0.7 \cdot 0.7 \cdot 0.7=125 \cdot 0.7^3&=42.875 \end{align*}

Since we multiply by 0.7 to find the 70% that is left after each hour, the decay factor is 0.7. The function will be:

\begin{align*}f(t)=125(0.7)^t.\end{align*}

We also found that after three hours the amount of the drug in the bloodstream will be 42.875 milligrams.

### Review

- Define exponential decay.
- What is true about “b” in an exponential decay function?
- Suppose \begin{align*}f(x)=a(b)^x\end{align*}. What is \begin{align*}f(0)\end{align*}? What does this mean in terms of the \begin{align*}y-\end{align*}intercept of an exponential function?

Graph the following exponential decay functions.

- \begin{align*}y=\frac{1}{5}^x\end{align*}
- \begin{align*}y=4 \cdot \left(\frac{2}{3}\right)^x\end{align*}
- \begin{align*}y=3^{-x}\end{align*}
- \begin{align*}y=\frac{3}{4} \cdot 6^{-x}\end{align*}

- The percentage of light visible at \begin{align*}d\end{align*} meters is given by the function \begin{align*}V(d)=0.70^d\end{align*}.
- What is the growth factor?
- What is the initial value?
- Find the percentage of light visible at 65 meters.

- A person is infected by a certain bacterial infection. When he goes to the doctor, the population of bacteria is 2 million. The doctor prescribes an antibiotic that reduces the bacteria population to \begin{align*}\frac{1}{4}\end{align*}of its size each day.
- Draw a graph of the size of the bacteria population against time in days.
- Find the formula that gives the size of the bacteria population in terms of time.
- Find the size of the bacteria population ten days after the drug was first taken.
- Find the size of the bacteria population after two weeks (14 days).

**Mixed Review**

- The population of Kindly, USA is increasing at a rate of 2.14% each year. The population in the year 2010 is 14,578.
- Write an equation to model this situation.
- What would the population of Kindly be in the year 2015?
- When will the population be 45,000?

- The volume of a sphere is given by the formula \begin{align*}v=\frac{4}{3} \pi r^3\end{align*}. Find the volume of a sphere with a diameter of 11 inches.
- Simplify \begin{align*}\frac{6x^2}{14y^3} \cdot \frac{7y}{x^8} \cdot x^0 y\end{align*}.
- Simplify \begin{align*}3(x^2 y^3 x)^2\end{align*}.
- Rewrite in standard form: \begin{align*}y-16+x=-4x+6y+1\end{align*}.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 8.8.