Let's look at exponential growth and decay.
What does it mean when we say something grows exponentially?
Here is an example. Suppose the population of a certain town
doubles every year and in the first year, let's call it year 0,
the population is 100 people. So we would have this:
when t equals 0, the population is 100.
In year 1, the population will have doubled since it's doubling every year.
So when t equals 1, the population is 200 people.
In year 2, the population will double from the previous year.
So when t equals 2, the population is 400 people.
We'll continue this for the first 10 years, so that you can see
the pattern that develops when we have exponential growth.
When t equals 3, the population is going to double from 400,
and it will be 800 people. When t is 4, the population is going to double from 800,
and you will have 1,600 people in the town.
When t equals 5, the population doubles from 1,600.
You'll have 3,200 people in the town in the fifth year.
When t equals 6, the population will be 6,400.
When t equals 7, the population will be 12,800.
When t equals 8, the population will be 25,600.
When t equals 9, the population will be 51,200.
Finally when t equals 10, the population will be 102,400 people.
Notice that after 10 years, the population is over 100,000 people.
This is characteristic of exponential growth.
Exponential growth appears to start off slow but eventually becomes very rapid.
Now we'll do a similar example for exponential decay.
Lead 210 is a radioactive element with a half-life of 22 years.
This means that given a certain amount of lead 210,
every 22 years, half of it will decay.
So let's suppose that we start off with 80 grams of lead 210.
Then the first 88 years would look like this:
When t equals 0, the amount of lead 210 is 80 grams.
When t equals 22 years, the amount of lead 210 would have been cut in half,
and we would have 40 grams.
When t equals 44 years, the amount from the 22 year mark would be cut in half,
and we would have 20 grams.
When t equals 66 years, the previous amount from the 44 year mark
would be cut in half, and we would have 10 grams.
Finally, when t is 88 years, we would have an amount of 5 grams.
Both exponential growth and exponential decay can be modeled by this equation:
A of t equals A sub zero times e to the k t,
where t is the time,
k is the growth or decay rate, A sub zero is the amount when t equals zero,
and A of t is the amount at time t.
If k is positive, this model is exponential growth,
and if k is negative this model is exponential decay.
Now let's do some examples. For example number one, we have
the size P of a certain insect population at time t, where t is in days,
obeys the function P of t equals 50 times e to the 0.03 t.
For part a, we would like the determine the number of insects when t equals 0.
For this calculation, we simply plug in zero for t,
and get P of zero equals 50 e to the 0.03 times zero which is 50.
Now this is e to the zero which has a value of one,
so 50 times 1 which gives us 50.
At time zero, there are 50 insects in this population.
Also notice that this is the same number A sub zero in the formula.
If I need to know the number of insects at time zero,
all I simply needed to do was pull A sub zero from the formula.
For part b, what is the growth rate of the insect population?
This is simply the value of k. In this problem, k is 0.03.
If I want to write this as a percent, that would be 3%.
Now we'll look at the next part of this problem.
For part c, we would like to determine the population of insects after 5 days.
For this calculation, we will find P of 5,
which will be 50 times e to the 0.03 times 5.
For this calculation, I will first multiply 0.03 times 5
and get P of 5 equals 50 times e to the 0.15
On the calculator, I will put in e to the 0.15.
This is second LN on this calculator. Put in 0.15.
If your calculator opened parenthesis, go ahead and close them at this point.
For e to the 0.15, I got approximately 1.16.
Take this number and multiply it by 50.
We'll get, for the approximate number of insects, 58.09.
Now, since the insect population would be a whole number,
we'll say the approximate number of insects is 58.
Now we'll look at the next part of this problem.
For part d, we would like the determine
the number of days it will take for the insects to reach 400.
To solve this, put 400 in for P of t.
We'll have 400 equals 50 times e to the 0.03 t
To solve this exponential equation, divide both sides by 50.
We'll get 400 over 50 equals e to the 0.03 t.
Reducing 400 over 50, we'll have 8 equals e to the 0.03 t.
To get the 0.03 t down from the exponent position,
we will take the natural log of both sides,
giving us the natural log of 8 is the natural log of e to the 0.03 t.
Using the properties of logarithms,
I can now take the 0.03 t, and write it in the front of natural log e.
So natural log of 8 is 0.03 t times the natural log of e.
Since the natural log of e equals 1,
this simplifies to natural log of 8 equals 0.03 t.
To find t divide both sides by 0.03,
and we'll get the natural log of 8 divided by 0.03 equals t.
I'm going to go ahead and approximate this on the calculator
and get natural log 8 divided by 0.03.
So the approximate number of days is 69.31.
Rounding to the nearest tenth, we get t is approximately equal to 69.3 days.
Now we will look at the next part of this problem.
For part e, we would like to determine
the number of days that it will take the insect population to double.
Since the initial amount was 50, this means for it to double,
P of t would be 100. I'll write 100 equals 50 times e to the 0.03 t.
To solve this equation for t, first divide both sides by 50.
We'll have 100 over 50 equals e to the 0.03 t.
Reducing 100 over 50, we'll have 2 equals e to the 0.03 t.
To solve this for t we will take the natural log of both sides,
and this will give us the natural log of 2 is the natural log of e to the 0.03 t.
This will allow us to take the exponent and put it in the front of natural log e.
So we'll have natural log 2 is 0.03 t times natural log e.
Since the natural log of e equals 1,
this simplifies to natural log 2 is 0.03 t.
Dividing both sides by 0.03, we'll get
natural log 2 divided by 0.03 equals t.
We'll approximate this on the calculator,
and get natural log 2 divided by 0.03.
t is approximately 23.1 days.
Rounded to the nearest tenth, t is approximately 23.1 days.
Now we'll take a look at the next example.
For example 2, we have the population of a colony of mosquitos
obeys the law of uninhibited growth.
If there are 1000 mosquitos initially and 1300 after 1 day,
then what is the size of the population after 3 days?
Since this colony follows the model of uninhibited growth,
we can model it with the function A of t equals A sub zero e to the k t.
For this problem, we will have to find the growth rate, k.
For this, substitute 1300 for A of t,
1000 for A sub zero, the initial amount,
e to the k, and t is 1.
When t is 1, there are 1300 mosquitos.
To solve this for k, we will have to divide both sides of the equation by 1000.
This gives us 1.3 equals e to the k.
Since k times 1 is k, I'll just write e to the k.
Take the natural log of both sides,
and get the natural log of 1.3 equals the natural log of e to the k.
Using the properties of logarithms, we can put k in the front of natural log of e.
k times the natural log of e is the natural log of 1.3.
Since the natural log of e is 1,
we will get k is approximately equal to,
let's find out the approximation for the natural log of 1.3.
Natural log of 1.3 is 0.262364
and this is an approximate value.
0.262364 for the growth rate.
From here, we'll go ahead and plug in k into the function,
and get A of t is 1000 e to the 0.262364 times t.
To find the size of the colony after 3 days, substitute 3 for t,
so A of 3 is 1000 times e to the 0.262364 times 3.
We can approximate this with the calculator,
and go ahead and multiply the exponent first.
.262364 times 3 will give me
A of 3 equals 1000 times e to the 0.787092793.
Now go ahead and use the answer feature from my calculator,
and do 2nd, LN, 2nd, ANS to get e to that power.
Multiply the result by 1000.
A of 3 is approximately equal to 2197 mosquitos.
Now we'll take a look at the next example. Example 3:
Strontium 90 is a radioactive material that decays according to the function
A of t equals A sub zero times e to the negative .0244 t, where t is in years.
Assume 800 grams initially. Find the half-life.
The half-life is the time it takes for half of the material to decay.
So if we are starting with 800 grams, then we need to know how long it will be
until there are 400 grams. So it will be 400 for A of t.
To solve this equation for t, divide both sides by 800.
So we'll have 400 over 800 equals e to the negative 0.0244 t.
Reducing 400 over 800, we get
1/2 equals e to the negative 0.0244 t.
To solve for t, take the natural log of both sides.
Natural log of 1/2 is the natural log of e to the negative 0.0244 t.
Using the properties of logarithms,
I can take the exponent and put it in the front of natural log e,
and end up with natural log 1/2 equals negative 0.0244 t times the natural log of e.
Since the natural log of e equals 1,
this simplifies to natural log of 1/2 is negative 0.0244 t.
Dividing both sides by negative 0.0244,
we'll get natural log of 1/2 divided by negative 0.0244 equals t.
Using the calculator, we'll get the approximate time for t.
This will be the natural log of 1/2,
make sure that when you put in natural log of 1/2, you close the parenthesis
before you type in divided by negative 0.0244.
Rounded to the nearest tenth,
we have that t is approximately equal to 28.4 years.
Suppose we did not know there were 800 grams initially.
We can still solve this problem.
We can substitute for A sub zero 1 unit and then for A of t, we would have 1/2 unit.
Since half-life is the amount of time it takes for the material to decay half way.
Substituting 1 for A sub zero and 1/2 for A of t gives me the same equation
that we have at this point in the solution
The solution from here would be the same for both problems.
Practice working these problems until you feel comfortable setting up and solving them.