8b)++Exponential+models+and+equations

Exponential models are commonly used to represent the growth or deprecation of things over a period of time. It is most commonly used to study changes in population, bacteria, electricity, temperature and credit payments.

__The equation that is used for such is:__ a = the initial value b = the base that is always positive x = the period of time

When b is greater than one, the model will display a growth. When the b is less than one but greater than zero, the model will display decay. There are two functions that can be used to graph growth and decay of objects.

__For growth you may use this equation:__ a = the initial investment of money r = the rate of change x = the period of time in which the rate of change will occur

__For decay you may use the equation:__ a = the initial investment of money r = the rate of change x = the period of time in which the rate of change will occur
 * NOTICE THAT THE ONLY DIFFERENCE BETWEEN THE TWO IS THAT ONE CONTAINS A SUM OF 1 AND r AND THE OTHER A DIFFERENCE.*

Step By Step Instructions on Solving Exponential Models and Equations for Growth: Example Question: The town of Nowheresville has a population of 340,000 and it increases 2 percent every year, how large will the population be in 50 years? 1) First we have to insert our values into the equation, and of course, remember to convert our rate of change to a decimal. a= 360,000 r= .07 x= 50 years

2) Next you have to add the percent of change to 1

3) Then raise the result to the power of 50 (the amount of time)

4) Finally multiply the initial amount by the result

At the end of 50 years their will be a population of 10,604,520 in Nowheresville.

Step By Step Instructions on Solving Exponential Models and Equations for Decay: Example Question: You buy a car for $13,000 and it depreciates at 12% per annually. What is its value after 7 years? 1) First we have to insert our values into the equation, and of course, remember to convert our rate of change to a decimal. a= $13,000 r= .12 x= 7 years The equation should look like this:

2) Next you have to subtract the rate of change from 1

3) Then raise the result to the power of 7 (the amount of time)

4) Finally multiply the result with your initial value

The result is $ 5,317

[|Problems Step By Step!]

Examples and Teaching [|Practice Problems]