Contributor: Jamie Hagler. Lesson ID: 13838
If your favorite candy bar gets 3% more expensive every year, how can you figure out what it will cost in 5 years? Find out with this lesson!
This is called exponential growth or decay.
When numbers increase or decrease rapidly, it is called exponential growth or decay.
Instead of the equation being linear like y = mx + b; it will be in the form y = ab^{x}.
Here's the setup of the equation:
For growth, we enter the number b + 1.
For decay, we enter the number 1 - b.
Notice that x is the exponent in this equation. That is why it is called an exponential function!
Let's try one!
The initial price of a boat is $25,000. It depreciates (decays, loses value) by approximately 10% every year.
$25,000 is the initial amount, so it will go in the place of a. It decays 10%, so we will enter 1 - 0.10 or 0.90.
The equation will look like this:
y = 25000(0.90)^{x}
There are several reasons; however, the main one is so we can predict the value of the boat each year.
Let's create a table to show the value each year. We will substitute the number of years into the place of the exponent x:
x | y = 25000(0.9)^{x} | y | ||
0 | 25000*(0.9)^{0} | |||
1 | 25000*(0.9)^{1} | |||
2 | 25000*(0.9)^{2} | |||
3 | 25000*(0.9)^{3} | |||
4 | 25000*(0.9)^{4} | |||
5 | 25000*(0.9)^{5} | |||
6 | 25000*(0.9)^{6} |
Then, we solve. These results will be our y values.
We can also write the coordinates of (x, y) so we can see these points on a graph:
x | y = 25000(0.9)^{x} | y | (x, y) | |
0 | 25000*(0.9)^{0} | 25000 | (0, 25000) | |
1 | 25000*(0.9)^{1} | 22500 | (1, 22500) | |
2 | 25000*(0.9)^{2} | 20250 | (2, 20250) | |
3 | 25000*(0.9)^{3} | 18225 | (3, 18225) | |
4 | 25000*(0.9)^{4} | 16402.5 | (4, 16402.03) | |
5 | 25000*(0.9)^{5} | 14762.25 | (5, 14762.03) | |
6 | 25000*(0.9)^{6} | 13286.03 | (6, 13286.03) |
If we plot these on a graph, we will see the exponential data is not linear, but makes a curve:
Let's try writing another exponential equation:
Stuart made an initial deposit of $5,000 in a savings account earning a yearly compounded interest rate of 3%. Assuming:
Let's put the numbers into the equation y = ab^{x}:
$5,000 is the initial amount, so it will go in the place of a.
The money earns interest of 3%, so we will enter 0.03 + 1 or 1.03.
The equation will look like this:
y = 5000(1.03)^{x}
Now that you have a handle on exponential growth and decay, let's move on to the Got It? section.
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