Suppose that 1000 people visited an online auction website during its first month in existence and that the total number of visitors to the auction site is tripling every month. Could you write a function to represent this situation? How many total visitors will the auction site have had after 9 months?

### Exponential Growth

Previously, we have seen the variable only used as the base in an exponential expression. In exponential functions, the exponent is the variable and the base is a constant.

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 growth situations, the growth factor must be greater than one.

\begin{align*}b>1\end{align*}

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

A colony of bacteria has a population of 3,000 at noon on Sunday. During the next week, the colony’s population doubles every day. What is the population of the bacteria colony at noon on Saturday?

Make a table of values and calculate the population each day.

Day |
0 (Sun) |
1 (Mon) |
2 (Tues) |
3 (Wed) |
4 (Thurs) |
5 (Fri) |
6 (Sat) |
---|---|---|---|---|---|---|---|

Population (thousands) | 3 | 6 | 12 | 24 | 48 | 96 | 192 |

To get the population of bacteria for the next day we multiply the current day’s population by 2 because it doubles every day. If we define \begin{align*}x\end{align*} as the number of days since Sunday at noon, then we can write the following: \begin{align*}P= 3 \cdot 2^x\end{align*}. This is a formula that we can use to calculate the population on any day. For instance, the population on Saturday at noon will be \begin{align*}P = 3 \cdot 2^6=3 \cdot 64 = 192\end{align*} thousand bacteria. We use \begin{align*}x=6\end{align*}, since Saturday at noon is six days after Sunday at noon.

#### Graphing Exponential Functions

Graphs of exponential growth functions show you how quickly the values of the functions get very large.

#### Let's use tables of values to complete the following problems:

- Graph \begin{align*}y=2^x\end{align*}.

Make a table of values that includes both negative and positive values of \begin{align*}x\end{align*}. Substitute these values for \begin{align*}x\end{align*} to get the value for the \begin{align*}y\end{align*} variable.

\begin{align*}x\end{align*} | \begin{align*}y\end{align*} |
---|---|

–3 | \begin{align*}\frac{1}{8}\end{align*} |

–2 | \begin{align*}\frac{1}{4}\end{align*} |

–1 | \begin{align*}\frac{1}{2}\end{align*} |

0 | 1 |

1 | 2 |

2 | 4 |

3 | 8 |

Plot the points on the coordinate axes to get the graph below. Exponential functions always have this basic shape: They start very small and then once they start growing, they grow faster and faster, and soon they become huge.

- In the last problem, we produced a graph for \begin{align*}y=2^x\end{align*}. Compare that graph with the graph of \begin{align*}y = 3 \cdot 2^x\end{align*}.

\begin{align*}x\end{align*} | \begin{align*}y\end{align*} |
---|---|

–2 | \begin{align*}3 \cdot 2^{-2} = 3 \cdot \frac{1}{2^2} =\frac{3}{4}\end{align*} |

–1 | \begin{align*}3 \cdot 2^{-1} = 3 \cdot \frac{1}{2^1} = \frac{3}{2}\end{align*} |

0 | \begin{align*}3 \cdot 2^0 = 3\end{align*} |

1 | \begin{align*}3 \cdot 2^1 = 6\end{align*} |

2 | \begin{align*}3 \cdot 2^2 = 3 \cdot 4 = 12\end{align*} |

3 | \begin{align*}3 \cdot 2^3 = 3 \cdot 8 = 24\end{align*} |

We can see that the function \begin{align*}y=3 \cdot 2^x\end{align*} is bigger than the function \begin{align*}y=2^x\end{align*}. In both functions, the value of \begin{align*}y\end{align*} doubles every time \begin{align*}x\end{align*} increases by one. However, \begin{align*}y=3 \cdot 2^x\end{align*} starts with a value of 3, while \begin{align*}y=2^x\end{align*} starts with a value of 1, so it makes sense that \begin{align*}y=3 \cdot 2^x\end{align*} would be bigger.

The shape of the exponential graph changes if the constants change. The curve can become steeper or shallower.

### Examples

#### Example 1

Earlier, you were told that 1000 people visited an online auction website during its first month in existence and that the total number of visitors to the auction site is tripling every month. What function represents this situation? How many total visitors will the auction site have had after 9 months?

To write the function, it is helpful to write out a table of values where \begin{align*}x\end{align*} is the month since the website opened and \begin{align*}y\end{align*} is the number of visitors

\begin{align*}x\end{align*} | \begin{align*}y\end{align*} |
---|---|

1 | 1000 |

2 | 3000 |

3 | 9000 |

4 | 27000 |

5 | 81000 |

6 | 243000 |

7 | 729000 |

8 | 2187000 |

9 | 6561000 |

Notice that the initial value is 1000 and the growth factor is 3. Therefore, you would expect the exponential function that represents this situation to be:

\begin{align*}y = 1000 \cdot 3^x\end{align*}

However, if you plug in 1 to this equation, you will get 3000 instead of 1000. The exponent value is 1 above what it should be. Therefore, you need to decrease the exponent by 1 and the exponential function that represents this situation is:

\begin{align*}y = 1000 \cdot 3^{x-1}\end{align*}

If you plug in 1 to this equation, you will get 1000 as necessary. You should test the other numbers from the table in this equation to verify that the function is correct.

To find how many total visitors that the website would have had after 9 months, add all the values of \begin{align*}y\end{align*} in the table from above:

\begin{align*}1000 + 3000 + 9000 + 27000 + 81000 + 243000 + 729000 + 2187000 + 6561000 = 9841000\end{align*}In the first 9 months, the website had 9,841,000 total visitors.

#### Example 2

A population of 500 *E. Coli* organisms doubles every fifteen minutes. Write a function expressing the population size as a function of hours.

Since there are four 15 minute periods in an hour, this means the population will double 4 times in an hour. Doubling twice is the same thing as quadrupling since:

\begin{align*}1\cdot 2 \cdot 2=1\cdot 2^2=1 \cdot 4=4.\end{align*}

This means that doubling 4 times can be calculated as \begin{align*}2^4=16\end{align*}. So the population is 16 times as big every hour. With an initial population size of 500, the function is:

\begin{align*}f(x)=500\cdot 8^x\end{align*}

where \begin{align*}x\end{align*} is in hours and \begin{align*}f(x)\end{align*} is the number of organisms after \begin{align*}x\end{align*} hours.

### Review

- What is the general form for an exponential equation? What do the variables represent?
- How is an exponential growth equation different from a linear equation?
- What is true about the growth factor of an exponential equation?
- True or false? An exponential growth function has the following form: \begin{align*}f(x)=a(b)^x\end{align*}, where \begin{align*}a>1\end{align*} and \begin{align*}b<1\end{align*}?
- What is the \begin{align*}y-\end{align*}intercept of all exponential growth functions?

Graph the following exponential functions by making a table of values.

- \begin{align*}y=3^x\end{align*}
- \begin{align*}y=2^x\end{align*}
- \begin{align*}y=5 \cdot 3^x\end{align*}
- \begin{align*}y=\frac{1}{2} \cdot 4^x\end{align*}
- \begin{align*}f(x)=\frac{1}{3} \cdot 7^x\end{align*}
- \begin{align*}f(x)=2 \cdot 3^x\end{align*}
- \begin{align*}y=40 \cdot 4^x\end{align*}
- \begin{align*}y=3 \cdot 10^x\end{align*}

Solve the following problems involving exponential growth.

- A chain letter is sent out to 10 people telling everyone to make 10 copies of the letter and send each one to a new person. Assume that everyone who receives the letter sends it to 10 new people and that it takes a week for each cycle. How many people receive the letter in the sixth week?
- Nadia received $200 for her \begin{align*}10^{th}\end{align*} birthday. If she saves it in a bank with a 7.5% interest rate compounded yearly, how much money will she have in the bank by her \begin{align*}21^{st}\end{align*} birthday?

**Mixed Review**

- Suppose a letter is randomly chosen from the alphabet. What is the probability the letter chosen is \begin{align*}M, K\end{align*}, or \begin{align*}L\end{align*}?
- Evaluate \begin{align*}t^4 \cdot t^\frac{1}{2}\end{align*} when \begin{align*}t=9\end{align*}.
- Simplify \begin{align*}28-(x-16)\end{align*}.
- Graph \begin{align*}y-1=\frac{1}{3} (x+6)\end{align*}.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 8.7.