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.

Example 15: Convert the exponential equation

to an equivalent equation with base e.

Solution and Explanations:

Let's make a few observations before we begin. Since the base in the equation

is , and is less than 1, the value of f(t) will get smaller as the value of t gets larger.

When t = 0 at the start of study, f(t) = 50,000; therefore, 50,000 is the starting amount. We are going to find a few points that satisfy this equation and use these points to find the new equation.

When t = 0, the value of f(t) is 50,000, and the corresponding point is (0, 50,000). At t = 1, the value of f(t) is

At t = 2, the value of f(t) is

At t = 3, the value of f(t) is

Let's start with a generic exponential equation with base e:
Substitute the point (0, 50,000) in the equation.

The equation can now be written
Substitute the point (1, 20,000) into the equation
Divide both sides of the equation by 50,000:
Solve for b by taking the natural logarithm of both sides of the equation .
Simplify the right side of the equation using the third rule of logarithms:
Simplify the left side of the equation: b = -0.916290731874, rounded to -0.91629.
Insert this value of b in the equation and the equation is simplified to
We know that the population is 8,000 after 2 minutes, so use these values to check the validity of the above equation. Substitute 2 for t in the right side of the above equation. If the answer is 8,000, then the model is correct.

rounded to 8,000. The model (equation) is correct.
We know from the original equation that after 3 seconds, the population is 3,200. Let's do a second check.