# 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 6: Town A had a population of 10,000 in 1900 and 20,000 in 1950. Town B had a population 60,000 in 1900 and 15,000 in 1950. In what year was the population in both towns the same?

Solution and Explanations:

In this problem, we will first have to find an exponential equation for each town. You could use any exponential equation of the form , where B is any positive number greater than 0, to solve this problem. For standardization sake, let's use the base e. Let the equation represent the population pattern for Town A, and let the equation represent the population pattern for Town B.

Town A:

The points on the graph of the function f(t) will be (specific time, and population at that time). Let's let t = 0 at the start of our study or 1900. Therefore, our first point is (0, 10,000). The second point corresponds to the year 1950 where t = 50 (1950 - 1900). The second point is (50, 20,000).
We will use these two points to find the value of a and the value of b in the equation

1.
Substitute the point (0, 10,000) into the equation :

2.
Substitute the value of a into the equation and we have

3.
Substitute the point (50, 20,000) into the equation and we have

4.
Divide both sides of the equation by 10,000:

5.
Solve for b by taking the natural logarithm of both sides of the equation .

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

7.
Divide both sides of the equation by 50 and simplify.

rounded to 0.0139.

8.
Insert this value of b in the above equation:

The f(t) stands for the population t years after the start of the study, the 10,000 is the population at the start of the study, the 0.0139 is the relative growth rate, and t equals the number of years since the start of the study in 1900.

The equation that best fits the population trend of Town A is

Town B:

The points on the graph of the function f(t) will be (specific time, and population at that time). Let's let t = 0 at the start of our study or 1900. Therefore, our first point is (0, 60,000). The second point corresponds to the year 1950 where t = 50 (1950 - 1900). The second point is (50, 15,000).

We will use these two points to find the value of c and the value of d in the equation

1.
Substitute the point (0, 60,000) into the equation :

2.
Substitute the value of c into the equation and we have

3.
Substitute the point (50, 15,000) into the equation and we have

4.
Divide both sides of the equation by 15,000:

5.
Solve for b by taking the natural logarithm of both sides of the equation

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

7.
Divide both sides of the equation by 50 and simplify.

rounded to -0.0277.

8.
Insert this value of d in the above equation:

The g(t) stands for the population t years after the start of the study, the 60,000 is the population at the start of the study, the -0.0277 is the relative growth rate, and t equals the number of years since the start of the study in 1900.

The equation that best fits the population trend of Town B

We are trying to determine in what year the population size was the same for these two town.. Since f(t) represents the size of the population of Town A and g(t) represents the size of the population of Town B, set f(t) equal to g(t) and solve for t.

When t = 43, or in the year 1943 (1900 + 43), the population of Town A and the population of Town B would be the same size. Let's check this number.

Town A: Insert t = 43 in the equation

Town B: Insert t = 43 in the equation

Close enough check. Remember the check will not be exact because we rounded the relative growth rates. Therefore, in 1943, the population of Town A and the population of Town B will be between 18,179 and 18,233.

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