In this lesson, we are going to talk about exponential functions. Previously, we spent quite a while looking at functions that are based around a variable raised to a number. But what if we took that idea and flipped it? We could consider functions that are a number raised to a variable, where we have some base number that has a variable as its exponent. These functions are called exponential functions, and we will explore them in this lesson. Make sure that you have a strong grasp on how exponents work before watching this.
where x is any real number and a is a real number such that a ≠ 1 and a > 0. We call a the base. [The base is the thing being raised to some exponent.]
From the previous lesson, Understanding Exponents, we can compute the value of a given base raised to any exponent. In practice, we can find these expressions (or a very good approximation) by using a calculator. Any scientific or graphing calculator
can do such calculations.
Exponential functions grow really, really, REALLY fast. There's a fictional story in the lesson to get across just how incredibly fast these things grow. Multiplying by a factor repeatedly can get to extremely
large values in a short period of time.
Compound interest is a form of interest on an investment where the interest gained off the principal (the amount of money initially invested) also gains interest. Thus, compound interest returns more and more money the longer the investment
is left in.
We can describe the amount of money, A, in a compound interest account with an exponential function:
A(t) = P
P is the principal in the account-the amount originally placed in the account,
r is the annual rate of interest (given as a decimal: 6% ⇒ 0.06),
n is the number of times a year that the interest compounds (n=1⇒ yearly; n=4⇒ quarterly; n=12 ⇒ monthly; n=365 ⇒ daily),
t is the number of years elapsed.
In the above equation, we can see that the more often the interest compounds, the more money we make. This motivates us to create a new number: the natural base, e. (See the video for more information on how we find e.) The number e goes on forever:
e = 2. 718 281 828 459 045 235 360 …
Like π, the number e is an irrational number: its decimal expansion continues forever, never repeating. Also like π, the number e is deeply connected to some fundamental things in math and the nature of the universe. Just as π is fundamentally
connected to how circles work, the number e is fundamentally connected to how things that are continuously growing work.
One application of e is to see how an account would grow if it was being compounded every single instant- being continuously compounded:
A(t) = Pert,
P is the principal (starting amount),
r is the annual rate of interest (given as a decimal),
t is the number of years elapsed.
The above equation can also be used for a wide variety of things that grow (or decay) continuously. You'll see it show up a lot in math and science.
We can see exponential decay if 0 < a < 1 in f(x) = ax. The exponential function will quickly become very small as it repeatedly "loses" its value because of the fraction (a) being compounded over and over.
Find the value of e3 to three decimal places.
The symbol e represents a number in the same way that π does. Also, just like π, e is irrational: its decimal expansion never ends and never repeats.
e = 2. 718 281 828…
Clearly, we can't use all the digits of e because there are infinitely many of them. However, the more of them we use, the more accurate our calculation will be. Since we need three decimal places of accuracy, let's begin by using at least six decimals from e: 2. 718 281. [This won't be perfectly accurate, but it will be accurate enough to find e3 to three decimal places. If we want more accuracy, we can use more decimals from e.]
Plug the approximation into a calculator:
e3 ≈ 2.7182813 = 20.085 518 558 …
Thus, we round to three places, saying e3 ≈ 20.086.
[Notice that the above approximation is not perfect. As we use more decimals, we get a more accurate result. Using a massive number of digits, we could find that
e3 = 20.085 536 923 …
See how the highly accurate value for e3 above starts to give different values after the fourth digit. While we can get good accuracy with a modest approximation of e, if we really need a massive amount of accuracy, we have to start with a lot of accuracy in the value we use for e.]
[If your answer is close, but off by the last couple digits, you probably need to use more digits of e. Look at the steps for a detailed explanation.]
Complete the table below for f(x) = 3x.
To fill in the table, plug in each value to the function.
For 3−1, remember that raising anything to the negative exponent causes it to "flip" to its reciprocal.
For 30, remember that raising anything to the 0 causes it to become 1.
As you fill out the table, notice how very quickly the function grows. Exponential functions become massively huge very, very quickly.
The function f(x) is an exponential function. Using the table of values below, state the function f(x).
An exponential function is a function of the form f(x) = ax, where a is some base. Thus, to answer the question, we need to figure out what a is.
The fact that f(0)=1 is not much help to us. As long as the function is strictly in the form ax, then any value of a will still produce 1. This is because raising any number to the 0 causes it to become 1. So we can't get much useful information from the first entry in the table.
The second and third entries are much more useful, though. We know f(2) = a2 = 22.09 and f(3) = a3 = 103.823.
Notice that a2 ·a = a3. Thus,
a2 ·a = a3 ⇒ 22.09 ·a = 103.823 ⇒ a = 4.7
Finally, now that we know a=4.7, we can double-check our answer. Currently, we have f(x) = 4.7x. From the table, we know f(4) should be 487.9681. Check using a calculator that 4.74 gives the same value. Indeed, it does, so our answer is correct.
f(x) = 4.7x
Graph the function f(x) = 4x−2−17.
Like graphing any function, we can always use brute force: calculate and plot a bunch of points, then draw a graph from those points. Approaching it that way, we might create a table such as the below:
Alternatively, we could examine the structure of the function. Notice that f(x) = 4x−2−17 is similar to the expression 4x. From looking at graphs of other exponential functions, it might be very easy for us to graph 4x: it would be 1 at x=0, 4 at x=1, and shoot up rapidly as we go farther to the right. If we examine 4x as it goes to the left, it would approach the x-axis asymptotically.
Now notice that 4x−2 would have the same graph as 4x, except it would be shifted 2 units to the right. Next, 4x−2−17 would be shifted a further 17 units down. [If you're not sure how to see this, check out the lesson on function transformations.] Thus, f(x) = 4x−2 −17 will have the same shape as 4x, but shifted 2 units right and 17 units down.
Either way that we approach understanding the shape of f(x), we now need to graph it. Notice that f(x) moves very quickly vertically for a small amount horizontally. Thus, we might want to choose graphing axes that are very tall for a small width.
Graph the function g(x) = ( [2/3] )x.
Because the base of the exponent is smaller than 1, the value of the function will become very small for positive x. As the fractions multiply each other more and more, the result will be tiny.
On the other hand, when we have a negative x, the negative exponent will cause the fraction to flip, and thus allow the function to grow.
To help us see how the function graphs, calculate various points:
Plot points and connect with curves. Notice that as x moves very far to the right, the function will asymptotically approach a value of 0. On the other hand, as x moves to the left, the function will grow very rapidly.
A bank account is opened and $10 000 is placed in the account. If the account has an interest rate of 4.7%, how much money is in the account after 25 years if the account compounds annually?
What if it had been compounded semiannually? Daily?
We can describe the amount of money, A, in such an account with an exponential function:
A(t) = P
where P is the principal in the account, r is the interest rate (as a decimal), n is the number of times the account compounds per year, and t is the number of years elapsed.
For this problem, we have
P=10 000, r = 0.047, t = 25.
The only thing left is how many times the account compounds per year, and the question asks us to figure that out for three different compounding schedules.
Those compounding schedules are as follows: annually-once a year ⇒ n=1; semiannually-twice a year ⇒ n=2; daily-every day, so 365 times in a year ⇒ n=365.
Use this along with the information above to figure out how much money is in the account for each situation.
Sally is currently 45 years old and is planning to retire at age 65. She has $25 000 in savings that she is going to put into an investment that has an interest rate of 7% and compounds continuously. How much money will she have when she retires?
How much money would she have at retirement if she had made the same investment, but had instead done it at age 35? What about age 25?
If something is being compounded continuously, the amount (A) of money after t years is
A(t) = Pert,
where P is the principal, r is the interest rate (in decimal), and e is the natural base.
The principal investment is P=25 000 and the rate is r=0.07. If she puts the money into the investment at age 45, then retires at age 65, that means the investment has t=20 years to mature. Thus, its value is
25000 e0.07 ·20 = 25000 e1.4 = 101 380.00
For the other ages, we just need to figure out new values for t. If she invests at age 35, then that means t=30 years until retirement:
25000 e0.07 ·30 = 25000 e2.1 = 204 154.25
If she makes the investment at 25, then she has t=40 years:
25000 e0.07 ·40 = 25000 e2.8 = 411 116.17
[Notice the massive increase in the value of the investment. Because compounding investments are based on exponential functions, the earlier the investment can be made, the greater its final value will wind up being.]
Age 45: $101 380.00, Age 35: $204 154.25, Age 25: $411 116.17
Bryan buys a new car for $26 650. Because of depreciation, the value of the car goes down with a time. The value of the car after t years is
p(t) = 26 650 (0.87)t.
What is the value of the car after one year of use? If Bryan sells the car after using it for a total of seven years, what will have been his net cost in owning the car? [Hint: The net cost is the original cost of the car, minus the amount he gets when selling the car.]
We have a formula to determine the value of the car after t years of use. To find the value of the car after one year of use, simply plug in t=1:
p(1) = 26650 (0.87)1 = 23 185.50
To find his net cost upon selling the car at seven years, we first need to know how much he sells the car for. Figure out the value of the car after seven (t=7) years:
p(7) = 26650 (0.87)7 = 10 053.84
Now that we know Bryan is able to sell the car for $10 053.84, we can calculate his net cost in owning the car. While it cost him $26 650 to purchase the car in the first place, he ultimately managed to sell it for some money, thus lowering the net cost of owning the car. To find the net cost, simply subtract the selling price from the original cost:
26650−10053.84 = 16 596.16
Value of car after one year: $23 185.50, Net cost of owning for seven years: $16 596.16
A population of bacteria doubles every 26 minutes. If the population starts with 50 cells, how many will there be in 12 hours? (Assume that none of the cells die off.)
Notice that we are told the population doubles after a certain time interval. This is equivalent to multiplying the starting population by 2 after a time interval passes. After the next time interval, it will multiply by 2 again, and so on.
Thus, if we say the number of time intervals is n, and the starting population is P, the amount A of bacteria will be
A = P·2n.
Now we need to figure out how many time intervals have elapsed. We know that one interval occurs every 26 minutes. The total time is 12 hours, which is 720 minutes. Thus, the number of intervals that have occurred is
We can plug that in to the above to figure out the amount of bacteria after 12 hours:
A = 50 ·2[720/26] = 10 843 894 130
10 843 894 130 bacteria
The radioactive isotope Carbon-14 has a half-life of 5 730 years. (Half-life is the time it takes for [1/2] of the isotope to decay and break down.) If we start with a quantity of 2.3 μg (micrograms) of C-14, how much will have not decayed after 30 000 years?
Notice that we are told the quantity halves after a certain time interval. This is equivalent to multiplying the starting amount by [1/2] after a time interval passes. After the next time interval passes, it will multiply by [1/2] again, and so on.
Thus, if we say the number of time intervals is n, and the starting quantity is P, the amount A of quantity in the end will be
A = P·
Now we need to figure out how many time intervals have elapsed. We know that one interval occurs every 5 730 years. The total time is 30 000 years. Thus, the number of intervals that have occurred is
We can plug that in to the above to figure out the quantity of not decayed C-14 after the passage of 30 000 years:
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.