Here is a standard textbook question: find an exponential function y = a*b^x whose graph passes through the points (1,6) and (3,24). The standard presentation produces a formula that is indeed correct, but the process used to arrive at it is not very intuitive, and does not give students a great deal of insight into the nature of exponential functions. But there is a better way! I’ll develop some ideas below, and I hope you’ll agree.
I love to use this 15-second video clip when exploring exponential growth with students:
Can we write a formula to describe the growth of the bacteria in the video? Assuming we start with just one bacterium, how many bacteria will there be in, say, 5 hours? And how do we model other, similar situations using exponential functions? Let’s explore.
I’ll stick with bacteria growth for the sake of giving students something concrete to keep in their minds as we do the math, but let’s allow the growth rates to be whatever we want (not necessarily “doubling” all the time).
At 12:00 PM a scientist observes 20 million bacteria in a petri dish. At 1:00 PM, she observes 60 million bacteria. How many bacteria can she expect to see by, say, 6:00 PM?
Solution: The boring version of this question is this: find an exp. function that contains the points (0,20) and (1,60). Here we start timing at 12:00 PM, so t = 0 at that time. We can figure out without a great deal of fuss that the answer must be y(t) = 20*3^t. Students should check using mental math that the two given points satisfy this equation. Two important observations: we had initially 20 million bacteria, and the number 20 is proudly displayed in the equation. Also, we can see that the number of bacteria tripled in one hour, and again the number 3 is on display in this equation. All well and good so far, but let’s play with this scenario a bit!
A scientist observes 20 million bacteria at 4:00 PM. By 5:00 PM, there are 60 million bacteria. Find a function that models these data.
Solution: Now we want the function to include (4, 20) and (5,60). Let’s be a little bit clever and *set our clocks back.* If instead of using t we use t – 4, we can instantly see that this function does the trick: y(t) = 20*3^(t-4). Again, students can use mental math to check that the two given points satisfy this equation. And students should still observe that the number of bacteria tripled in one hour. The equation displays the amount of bacteria at t = 4, the fact that the number triples in one hour, and the fact that we have set our clocks back 4 hours. That wasn’t too bad! Let’s make the problem a touch harder.
A scientist observes 20 million bacteria at 12:00 PM. By 5:00 PM, there are 60 million bacteria. Find a function that models these data.
Solution: This time we need the function to include (0, 20) and (5, 60). And now we can use a second clever trick: let’s *compress time!* This time it took 5 hours to triple the number of bacteria. But instead of thinking of this as 5 of something, why not instead think of it as 1 of something? We can do this by organizing time into 5-hour intervals. How is this accomplished? Just use t/5 instead of t! Surely the two given points satisfy the equation f(t) = 20*3^(t/5), as students can check mentally. Once again we see from the equation the role that “20 million bacteria” play, the role that “tripling” plays, and as a bonus we see that we have compressed time by a factor of 5, treating 5 hours as 1 block of time!
Once students are comfortable setting the clock back, and are comfortable compressing time, it’s time to bring both features together at once!
A scientist observes 20 million bacteria at 4:00 PM. By 9:00 PM, there are 60 million bacteria. Find a function that models these data.
Solution: First we observe that it took 5 hours for the number of bacteria to triple. Now we seek the function that includes (4, 20) and (9, 60). Clever move #1: set the clock back so that 4:00 PM behaves like noon: use t – 4. Clever move #2: compress time so that 5 hours feel like one long block of time: use t/5. But wait, we are not using t, we’re using t-4! So make that (t-4)/5. With these simple tricks, we quickly arrive at the answer: y(t) = 20*3^(t-4)/5. It is still a matter of doing simple mental math to check that (4,20) and (9,60) satisfy this equation! And we can clearly see the role that the number 20 million plays (we knew there were that many at 4:00 PM), the role that tripling plays (the ratio of 60 million to 20 million), the fact that we delayed time by 4 hours (since we started at 4:00 PM), and the fact that we compressed time by a factor of 5 (since 9:00 PM is 5 hours after 4:00 PM). Brilliant!
A mathematician observes that y = 19 when t = 13, and that y = 25 when t = 23. He seeks an exponential function that fits these data, and he needs our help!
Solution: using the techniques described above, we have the answer in a jiffy! y(t) = 19*(25/19)^(t-13)/10. Check for yourself that this function includes (13,19) and (23,25). This function was created to make checking these two points as easy as possible!
Observations about the equation y(t) = 19*(25/19)^(t-13)/10:
- 19 is the y-value when t = 13, the “first observation.”
- 25/19 is the ratio of the two known y-values. This conveys something about the rate at which the y-values are growing.
- t – 13 is a convenient expression, since the first point has t = 13.
- (t – 13)/10 is also convenient, since the space between t = 13 and t = 23 is “10 hours.”
That’s it! Once students have enough practice, they can get their answers with little or no work, and with complete understanding!
- Set the clock back! Use a transformation of t into t – (something).
- Compress time! Change t into t / (something).
- Now you are ready to rock and roll! The math becomes easy once you do those two things. Answers pop out in about two or three lines of writing!
Thanks to Dr. James Tanton for inspiring this essay. You can see his treatment of this subject here.