# APPLICATIONS OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS

APPLICATIONS OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS

(Population Word Problems)

To solve an exponential or logarithmic word problem, convert the narrative to an equation and solve the equation. In this section, we will review population problems.

Problem 4: Find the equation to the base 50 that describes a population of cells that start with 5 cells and double every hour.

Solution and Explanations:

First record your observations by making a table with two columns: one column for the time and one column for the number of cells. The number of cells (population size) depends on the time. If you were to graph your findings, the points would be formed by (specific time, number of cells at the specific time). For example at t = 0, there are 5 cells, and the corresponding point is (0, 5). At t = 1, there are twice as many cells or 10 cells, and the corresponding point is (1, 10). At t = 2, there are 20 cells, and the corresponding point is (2, 20). At t = 3, there are 40 cells, and the corresponding point is (3, 40).

Let's start with the generic exponential function with a base equal to e.

where a is the size of the population (number of cells) at the beginning of the student, t equals time into the study, and f(t) is the population at time t.

1.
Using the point (0, 5), substitute 0 for t and 5 for f(t) in the equation .

2.
Rewrite the equation with a = 5.

3.
Using the point (1, 10), substitute 1 for t and 10 for f(t) in the equation .

4.
Divide both sides of the equation by 5.

5.
Take the natural logarithm of both sides of the equation .

6.
Simplify the right side of the equation using the third rule of logarithms.

rounded to 0.1772

7.
Substitute this value of b in the equation and we have

8.
Use the point (2, 20) to check our answer. Substitute 2 for t and determine whether this substitution will yield a value of 20 for f(t). If it does, the equation is correct.

9.
The check is not exact because we rounded the 0.177184 to 0.1772.

10.
Use the point (3, 40) as a second check. Substitute 3 for t and determine whether this substitution will yield a value of 40 for f(t). If it does, the equation is correct.

11.
The check is not exact because we rounded the 0.177184 to 0.1772.

If you would like to work another problem, click on Problem

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