Exponential Relations

Section 6.1 Exponential Relations

In Section 3.2 we learned different ways data can change. These included linear, quadratic, exponential and inverse versions of these. Here we will look at more details of exponential relations.

For exponential relations we know that the differences (additive rate) is a multiple of the value (Definition 3.2.12) and that the percent increase is constant (Definition 3.2.16). Now we will learn to work with these relations in applications.

Subsection 6.1.1 Comparing Growth Rates

In colloquial speech exponential is used to mean “very fast”. We are using a much more specific definition. Here we will see why we think of exponential as very fast. We will do this by comparing how fast exponential relations grow relative to linear, quadratic, and one new relation.

Consider the linear, quadratic, and exponential relations in Table 6.1.1, Table 6.1.2, and Table 6.1.3 respectively. Because the differences for a linear relation are always the same, eventually the values are bigger than the change (e.g., 5,8,11 are all bigger than 3). For both quadratic and exponential data the bigger they are the faster they grow. The growth is not the same however.

For the quadratic relation the first differences are getting bigger (at a constant rate), but eventually they are less than the value (e.g., 6,10,14 are less than 8,18, and 32 respectively) just like the linears. For exponential relations, however, the rate of growth of the values is a multiple of the values, that is the rate of growth increases as quickly as the values do (e.g., the values are doubling and so are the differences). Because of this exponential growth will always outpace linear and quadratic eventually.

Table 6.1.1. Linear Relation

$n$	0	1	2	3	4	5
$3n+2$	2	5	8	11	14	17
Differences		3	3	3	3	3

Table 6.1.2. Quadratic Relation

$n$	0	1	2	3	4	5
$2n^2$	0	2	8	18	32	50
Differences		2	6	10	14	18

Table 6.1.3. Exponential Relation

$n$	0	1	2	3	4	5
$3(2^n)$	3	6	12	24	48	96
Differences		3	6	12	24	48

We can also compare the ratios of consecutive terms (which can be thought of as percent increase) of the relations. Table 6.1.4 has the ratios for a linear relation. Notice that $\frac{8}{5} < \frac{5}{2}\text{.}$ This inequality is true of the other ratios. The ratios are becoming smaller; indeed they appear to be approaching 1. In Table 6.1.5, the ratios for an exponential relation are constant. This also shows us that exponential relations grow faster than linear relations.

Table 6.1.4. Linear Relation

$n$	0	1	2	3	4	5
$3n+2$	2	5	8	11	14	17
Ratios		$5/2$	$8/5$	$11/8$	$14/11$	$17/14$

Table 6.1.5. Exponential Relation

$n$	0	1	2	3	4	5
$3(2^n)$	3	6	12	24	48	96
Ratios		2	2	2	2	2

Subsection 6.1.2 Faster than Exponential

The previous section might suggest that exponential is the fastest relation. It is not. In Table 6.1.6 are the ratios for a relation known as factorial. Interestingly, the ratios giving the percent increase are linear. This is vaguely like quadratic relations where their differences are linear. As quadratic relations are faster growing than linear relations, so factorial relations are faster growing than exponential relations.

Table 6.1.6. Factorial Relation

$n$	0	1	2	3	4	5
$n!$	1	1	2	6	24	120
Ratios		1	2	3	4	5

Subsection 6.1.3 Applications

The bacteria lactobacilus acidophilus is known to grow exponentially in milk (part of making yogurt). Based on experiments a new generation of bacteria are formed every 70 minutes. For a while we can assume the original bacteria remain. Thus the population growth is as shown in Table 6.1.7. Note that at 70 minutes the model supposes all 4000 cells divide producing 4000 new cells. Thus the total population is $4000+4000=8000\text{.}$ This repeats every 70 minutes. Because the cells split into two cells, the population doubles (multiply by 2).

Table 6.1.7. Growth of Lactobacilus Acidophilus

Minutes	0	70	140	210
Population	4000	8000	16000	32000
Added		4000	8000	16000

We can express this model in mathematical notation. Let’s walk through how Table 6.1.7 was constructed. The initial amount (information we are given) is $P_0 = 4000\text{.}$ After 70 minutes these double: $P_1 = 4000(2) = 8000\text{.}$ After 140 minutes they double a second time: $P_2 = 4000(2)(2) = 16000\text{.}$ After 210 minutes they double a third time: $P_3 = 4000(2)(2)(2) = 32000\text{.}$

Generally if we know the number of minutes $t\text{,}$ then dividing by 70 tells us how many times cells have divided in two. We want to multiply the initial amount (4000 in this case) by two (double the population) for each of these. This gives us $P = 4000 \cdot 2^{t/70}\text{.}$

Example 6.1.8.

A new cat video is posted and 12 people view it. Every 4 days after they view it the number of people who see it triples. Write an equation to model the number of people who view it on a given day.

Solution.

We can calculate the first few days results. The 4 days after the video is posted, there will be 36 views. The eighth day, there will be 108 views. Each day we multiply the result by 3. Note that the tripling occurs every four days we need to divide the number of days by 4 to determine how many times it has tripled. We multiply the original twelve by 3 to triple it. Thus the number of people viewing the video is

\begin{equation*} v = 12(3^{d/4}) \end{equation*}

Checkpoint 6.1.9.

The bacteria bacillus megaterium grows exponentially. It forms a new generation every 25 minutes. Write an equation to model this grows supposing there are 30 g bacteria at the beginning.

Solution.

\begin{equation*} b = 30(2^{t/25}) \end{equation*}

Note that in general these types of exponential growth models look like

\begin{equation*} P = P_0 (r^t) \end{equation*}

where $P$ is the current population (or count), $P_0$ is the initial population (or count), and $r$ is the rate at which the population grows (e.g., doubles, triples).

To write the model (equation) we need to know an initial amount and the rate of increase or decrease.

Example 6.1.10.

Suppose we know that over a period of time a population of hares are growing exponentially. The initial population was 320 hares and they doubled every 200 days. Write the equation to model this population.

Solution.

The initial population is $P_0=320\text{.}$ Because the rate is doubling, we know that $r=2\text{.}$ If we want the variable to be in units of days, we need to adjust for the 200 day period. This is accomplished by using $t/200\text{.}$ Note this means 200 gives doubling once ($200/200=1$), and 400 gives doubling twice ($400/200=2$).

\begin{equation*} P=320(2^{t/200}) \end{equation*}

We learned in Subsection 3.2.2 that salaries increased by a fixed percent each year are exponential in nature. Now we can write a model for this and calculate results.

Example 6.1.11.

Tien’s initial salary was $52,429.33. He received a 5% raise each year. What should Tien’s salary be entering the sixth year?

Solution.

Because the raise is a 5% increase, the percent is $p=1.05\text{.}$ The model then is

\begin{equation*} S=\$52,429.33(1.05)^t \end{equation*}

where $t$ is the number of years since he was hired. Entering the sixth year would mean five raises. His salary would be

\begin{equation*} S=\$52,429.33(1.05)^5 = \$66,914.59. \end{equation*}

Example 6.1.12.

If Moses’ salary after six raises was $72,311.54, and he received a 4% raise each year. What was his initial salary?

Solution.

Because the raise is a 4% increase, the percent is $p=1.04\text{.}$ The model then is

\begin{equation*} S=S_0(1.04)^t \end{equation*}

where $t$ is the number of years since he was hired, and $S_0$ is the initial salary. We can now solve for the initial salary.

\begin{align*} \$72,311.54 \amp = S_0(1.04)^6.\\ \$72,311.54 \amp = S_0(1.2653).\\ \frac{\$72,311.54}{1.2653} \amp = \frac{S_0(1.2653)}{1.2653}.\\ \$57,149.72 \amp = S_0. \end{align*}

Note, because the salary would be rounded each year this might be off by a small amount, but not enough to matter for our curiosity.

Example 6.1.13.

We will calculate the percent increase given initial and final salaries. If Raven’s initial salary was $53,242.17, and her salary at the end of 7 years was $67,368.33, what was her annual percentage increase?

Solution.

The end of seven years means there have been six raises. From the two data points we know

\begin{align*} 67368.33 & = 53242.17(r^6)\\ \frac{67368.33}{53242.17} & = r^6\\ \sqrt[6]{\frac{67368.33}{53242.17}} & = \sqrt[6]{r^6}\\ 1.04 & = r \end{align*}

Thus her annual percentage increase was 4%. The full model is

\begin{equation*} S = 53242.17(1.04)^t \end{equation*}

where $t$ is time in years.

Subsection 6.1.4 Interpretation of Exponentials

There is more than one way to interpret the base (e.g., the 7 in $P=5(7^n)$). When it is an integer like 2 ($P=5(2^n)$) or 3 ($P=7(3^n)$) it tells us that the data is doubling or tripling. When the base looks like 1.053 ($P=100(1.053)^n$) it tells us the percent increase. In this example this represents a 5.3% increase. Recall from Subsection 2.1.2 that a 3% decrease is represented by $100\%-3\%=97\%\text{.}$ In an exponential model this would look like $P=100(0.97)^n\text{.}$ Generally interpreted as a percent the exponential model is

\begin{equation*} P=P_0(r)^n=P_0(1 \pm p)^n. \end{equation*}

Example 6.1.14.

Suppose $P=100(1.032)^t$ where $t$ is the number of years. What is the annual percent increase?

Solution.

We know that

\begin{align*} 1.032 \amp = 1+p\\ 1.032-1 \amp = p\\ 0.032 \amp = p \end{align*}

so this is a 3.2% increase.

Example 6.1.15.

Suppose $P=100(0.88)^t$ where $t$ is the number of years. What is the annual percent decrease?

Solution.

We know that

\begin{align*} 0.88 \amp = 1-p\\ 0.88-1 \amp = -p\\ -0.12 \amp = -p\\ 0.12 \amp = p \end{align*}

so this is a 12% decrease.

Example 6.1.16.

Suppose $P=230(1.03)^t\text{.}$ Is this increase or decrease and what is the percent increase/decrease?

Solution.

Because $1.03 > 1.00$ this is a percent increase.

\begin{align*} 1.03 \amp = 1 + p\\ 1.03-1 \amp = p\\ 0.03 \amp = p \end{align*}

Thus this is a 3% increase.

Even when the exponential model is expressed in terms of fractions, we can determine the percent increase or decrease.

Example 6.1.17.

In the exponential model $P=342\left(\frac{7}{5}\right)^n$ what is the percent increase or decrease?

Solution.

We convert the fraction $7/5$ to decimal: 1.40. Because it is bigger than one it is a percent increase. Note $1.40=1+0.40\text{,}$ so this is a 40% increase.

Example 6.1.18.

In the exponential model $P=342\left(\frac{39}{40}\right)^n$ what is the percent increase or decrease?

Solution.

We convert the fraction $39/40$ to decimal: 0.975. Because it is less than one it is a percent decrease. Note $1-0.975=0.025\text{,}$ so this is a 2.5% decrease.

A common base for exponentials in many scientific models is $e\text{.}$ We can work with this base using calculation devices which will have an $e^x$ button or an $\text{exp}(x)$ function.

Example 6.1.19.

What is the percent increase or decrease for the exponential model $P=27e^{0.02t}\text{?}$

Solution.

Just as with the fractions we convert to decimal. Note that we can also write the model $P=27(e^{0.02})^t\text{.}$ Do you see the difference? $e^{0.02} \approx 1.0202\text{.}$ Thus this is a percent increase of 2.02%.

Checkpoint 6.1.20.

In the model $P=173e^{0.05t}$ what is the percent increase or decrease (2 decimal places)?

It is a percent

Increase
Decrease

Answer 1.

$\text{Increase}$

Answer 2.

$5.13$

Solution.

We calculate $e^{0.05} \approx 1.051271096\text{.}$ This is greater than one so it is a percent increase. The percent is 5.13

\(n\)	0	1	2	3	4	5
\(3n+2\)	2	5	8	11	14	17
Ratios		\(5/2\)	\(8/5\)	\(11/8\)	\(14/11\)	\(17/14\)

Section 6.1 Exponential Relations

Subsection 6.1.1 Comparing Growth Rates

Subsection 6.1.2 Faster than Exponential

Subsection 6.1.3 Applications

Example 6.1.8.

Checkpoint 6.1.9.

Example 6.1.10.

Example 6.1.11.

Example 6.1.12.

Example 6.1.13.

Subsection 6.1.4 Interpretation of Exponentials

Example 6.1.14.

Example 6.1.15.

Example 6.1.16.

Example 6.1.17.

Example 6.1.18.

Example 6.1.19.

Checkpoint 6.1.20.

Exercises 6.1.5 Exercises

Exercise Group.

1. Contextless Practice.

2. Contextless Practice.

3. Contextless Practice.

4. Contextless Practice.

5. Contextless Practice.

6. Contextless Practice.

7. Contextless Practice.

8. Faux Application.

9. Application.

10. Application.

11. Application.

12. Application.

13. Application.

14. Application.

15. Application.

16. Application.

17. Application.

18. Application.

Exercise Group.

19. Contextless Practice.

20. Contextless Practice.

21. Contextless Practice.

22. Contextless Practice.

23. Contextless Practice.

24. Contextless Practice.

25. Contextless Practice.

26. Interpreting (Contextless).

27. Interpreting (Contextless).

28. Interpreting (Contextless).

29. Interpreting (Contextless).