Applications of Exponential Functions

There are important applications of exponential functions in everyday life. The most important applications are related to population growth, exponential decline, and compound interest. These situations can be easily modeled with exponential functions. It is possible to predict future scenarios with the knowledge of certain current parameters.

ALGEBRA

Relevant for

Learning about some applications of the exponential function.

See applications

ALGEBRA

Relevant for

Learning about some applications of the exponential function.

See applications

1. Population growth

In some cases, scientists start with a certain number of bacteria or animals and watch their population change. For example, if the population is doubling every 7 days, this can be modeled by an exponential function.

The general formula used to represent population growth is $latex P (r, t, f) = P_{i} {{(1 + r)}^{\frac{t}{f}}}$, where $latex P_{i }$ represents the initial population, r is the population growth rate, t is the elapsed time, and f is the period over which the population grows by a rate of r.

The ratio of t to f is many times simplified into a single value that represents the number of compound cycles. Although this is the general formula, many of the population models use the number e and form the formula $latex P P_{i}{{e}^{kt}}$.

Population models can occur in two ways. One way is if we have an exponential function and we have to find estimates. The second way involves finding an exponential equation from the given information. We will look at examples both ways.

EXAMPLE 1

The population of a city is $latex P=350 000{{e}^{0.012t}}$, where $latex t=0$represents the population in the year 2020.

a) Find the population of the city in the year 2030.

Solution: To find the population in the year 2030, we need to use $latex t=10$ in the given equation:

$$P=350 000{{e}^{0.012t}}=350 000{{e}^{0.12}}=394 624$$

b) Find the population of the city in the year 2035.

Solution: To find the population in the year 2035, we need to use $latex t=15$:

$$P=350 000{{e}^{0.012t}}=350 000{{e}^{0.18}}=419 026$$

c) Find when the population will equal 450,000.

Solution: From the previous question, we know that the population in 2035 will be 419,026, therefore, it makes sense that the answer is greater than 2035. Remember that the P in the equation is the final population, which we have as 450,000. The equation that we have to solve to find the time is:

$latex 450 000=350 000{{e}^{0.012t}}$

Thus, we solve as follows:

$latex 450 000=350 000{{e}^{0.012t}}$

$$\frac{450 000}{350 000}={{e}^{0.012t}}$$

$$\ln\left( \frac{450 000}{350 000}\right)=\ln {{e}^{0.012t}}$$