+
Algebra I: Exponential Growth and Decay

Algebra I: Exponential Growth and Decay

Author: Jessica Yang
Description:

To gain an understanding of exponential growth and decay, and to learn how they are relevant in the real world.

We will explore the backbone of exponential growth and decay and their graphical interpretations.

(more)
See More

Try Our College Algebra Course. For FREE.

Sophia’s self-paced online courses are a great way to save time and money as you earn credits eligible for transfer to over 2,000 colleges and universities.*

Begin Free Trial
No credit card required

25 Sophia partners guarantee credit transfer.

221 Institutions have accepted or given pre-approval for credit transfer.

* The American Council on Education's College Credit Recommendation Service (ACE Credit®) has evaluated and recommended college credit for 20 of Sophia’s online courses. More than 2,000 colleges and universities consider ACE CREDIT recommendations in determining the applicability to their course and degree programs.

Tutorial

Math Wizard's Notes

Exponential growth and decay are rates. They represent the change in some quantity through time.

Exponential Growth

Exponential growth is any increase in a quantity P.

     P(t)= Poe^(kt)

Where Po is the initial quantity, t is time, k is a constant, P(t) is the quantity after time t, and e^x is the exponential function.

Exponential Decay

Exponential Decay is an decrease in a quantity P.

     P(t)= Poe^(-kt)

Where Po is the initial quantity, t is time, k is a constant, P(t) is the quantity after time t, and e^x is the exponential function.

Math Wizard's Notes

Exponential Growth and Decay Models

Po<k is a graphical interpretation of exponential decay.

Po>k is a graphical interpretation of exponential growth.

Po=k is a graphical interpretation of constant exponential growth/decay.

Math Wizard's Exercise

1. If a city has a population of 340 people, and if the population grows continuously at an annual rate of 2.3%, what will the population be in 6 years?

     We are given Po= 340 people and k= .023.

     P(t)= 340e^(.023t)

     When t= 6 years

     P(t)= 340e^(.023*6)= ~390 people

 

2. If China has a population of 420 tigers, and if the population doubles every 9 years, what will the population be in 7 years?

     We are given Po= 420 and t= 7 and P(9)= 840.

     A doubling time of 9 years means that 840= P(9)= Poe^(kt), and follows that 2=  e^(9k).

     Take the natural log of both sides to get ln2=9k.

     So k=(ln2)/9

     Hence, P(7)= 420e^(7k)= 420e^((7ln2)/9)= ~720 tigers