Exponential functions problem

In summary, the conversation discusses two types of exponential growth/decay problems. The first involves a cell losing 2% of its charge every day, while the second involves the population of a country increasing by 1.4% per year since 1981. Both equations follow the form of P(t) = P_0 (1 + r)^t, where r represents the rate of growth/decay. In the first equation, r is equal to -0.02, while in the second equation, r is equal to 0.014. The 1.014 constant in the second equation represents the growth rate by which the population increases each year, and was given in the problem.
  • #1
supernova1203
210
0

Homework Statement


In the 2 following problems they use the term in the brackets differently, in one case its a percentage and in the other case i have no idea where they get the number from, this is what i would like to find out

A cell loses 2% of its charge every day
C is total charge t is time(measured in days)

C(t)=100(0.98)t

so basically for each day we put a number in the exponent, for example if 4 days have passed then we write

C(t)=100(0.98)4

So here 100 represents the total charge, 0.98 represents the percentage of charge left after 1 day. This is one type of exponential growth/decay problem I see and is easy to solve.

This other type of problem is where i am having a bit of trouble:

The population of a country in 1981 was 24 million

P represents the population in millions and t represents the time in years

The following equation represents this model

P(t)=24(1.014)t
Here 24 represents the total population in 1981, 1.014 is the growth rate by which the population increases each year(in the millions) and the exponent is where we put in the number of years it has been since 1981

so for example if we use 2011 (1981+30=2011) that means to get the population for 2011
we do P(t)=24(1.014)30

which gives us the answer

P(t)= 24(1.517534768)
P(t)=36.4 million.

my question is where or how do they get the 1.014 constant which is used to calculate the population for every year relative to 1981? It is the exponential growth multiple isn't it? How did they get it?

Homework Equations



P(t)=24(1.014)t

The Attempt at a Solution


 
Physics news on Phys.org
  • #2
I usually think of exponential growth/decay models in terms of this equation:
[tex]P(t) = P_0 (1 + r)^t[/tex]
r is the rate of growth/decay. r > 0 would indicate a growth, and r < 0 would indicate a decay.

So for the 1st equation,
[tex]C(t) = 100(0.98)^t = 100(1 - 0.02)^t[/tex]
... since a cell loses 2% of its charge every day.

For the 2nd equation,
[tex]P(t) = 24(1.014)^t = 24(1 + 0.014)^t[/tex]
... so the rate of growth is 1.4% per year since 1981. This information should have been given in the problem somewhere.
 
  • #3
eumyang said:
I usually think of exponential growth/decay models in terms of this equation:
[tex]P(t) = P_0 (1 + r)^t[/tex]
r is the rate of growth/decay. r > 0 would indicate a growth, and r < 0 would indicate a decay.

So for the 1st equation,
[tex]C(t) = 100(0.98)^t = 100(1 - 0.02)^t[/tex]
... since a cell loses 2% of its charge every day.

For the 2nd equation,
[tex]P(t) = 24(1.014)^t = 24(1 + 0.014)^t[/tex]
... so the rate of growth is 1.4% per year since 1981. This information should have been given in the problem somewhere.

so the only difference between the first and the second equation is that the first equation rate(r) is not multiplied by 100 and the second one rate(r) is?
 
  • #4
supernova1203 said:
so the only difference between the first and the second equation is that the first equation rate(r) is not multiplied by 100 and the second one rate(r) is?

I don't know what you mean. r = -0.02 corresponds to a 2% loss rate. r = +0.014 corresponds to a 1.4% rate of growth. We're just converting to percents, and in both cases we multiply 100 to convert to percents. :confused:
 

What is an exponential function?

An exponential function is a mathematical function of the form f(x) = ab^x, where a and b are constants and x is the independent variable. It is characterized by a rapid increase or decrease in values as x changes, and the base b determines the rate of change.

How do I solve exponential function problems?

To solve an exponential function problem, you need to follow these steps:

  1. Identify the base and exponent in the function.
  2. Substitute the given values for the base and exponent into the function.
  3. Simplify the expression using the rules of exponents.
  4. Solve for the variable, if necessary.

What is the difference between exponential growth and decay?

Exponential growth is when the output of an exponential function increases as the input increases. It is represented by a positive base (b > 1) in the function. Exponential decay, on the other hand, is when the output decreases as the input increases. It is represented by a base between 0 and 1 (0 < b < 1) in the function.

What are some real-life applications of exponential functions?

Exponential functions can be used to model many real-life phenomena, such as population growth, compound interest, radioactive decay, and the spread of diseases. They are also commonly used in finance, biology, physics, and other fields to analyze and predict data.

What are some common mistakes to avoid when working with exponential functions?

Some common mistakes to avoid when working with exponential functions include forgetting to apply the rules of exponents, using the wrong base or exponent, and not simplifying the expression before solving for the variable. It is also important to pay attention to the units and make sure they are consistent throughout the problem.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
5
Views
2K
  • Precalculus Mathematics Homework Help
Replies
5
Views
2K
  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
12
Views
1K
  • Precalculus Mathematics Homework Help
Replies
9
Views
2K
Replies
3
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
6
Views
935
  • Precalculus Mathematics Homework Help
Replies
11
Views
2K
Replies
3
Views
2K
  • Precalculus Mathematics Homework Help
Replies
4
Views
708
Back
Top