Some Practice with Exponential Growth and Decay
Recall that the defining equation for growth or decay at a constant rate is
N  N0e kt
where N0 is the value of the growing or decaying quantity at t = 0 (the arbitrary time when you
start the problem), k is the growth or decay rate (how many % per year N grows or decays,
expressed as a decimal: + 0.05 for 5 % growth per year, – for decay), and t is time in whatever
units you choose (beware, however; if k is change per year, t is in years; if k is change per day, t
must be measured in days). Let's try an example. Suppose the inflation rate over the last few
decades has averaged about 6% per year (it's really low now, but was around 14% in the 1970's,
so we're taking an average). I can barely remember when a gallon of gas cost about $0.50,
sometime in the 1960's. Maybe my memory is foggy, so let's say a gallon of gas cost $0.30 in
1960. If inflation remained steady at 6% per year, what would it cost today? Let's change the
symbol from N to P (for price) and write the equation above as
P  P0ekt  $0.30 e0.06t
Today would be 40 years after 1960 (our starting time for the problem), so we use t = 40 and get
P  $0.30 e
0.06(40)
 $3.30
Wow! Maybe gas prices aren't so high after all! Obviously, something must be keeping gas
prices lower than we would predict based on the historic inflation rate. Or maybe we just got our
numbers wrong. Whatever the case, the lesson is clear. Exponential growth, even at fairly low
rates, will usually give a surprising result.
Now let's ask a different question. Assuming a gallon of gas cost $0.30 in 1960 and that
the price grows at 6% per year, in what year will you have to pay $20 per gallon? With P = $20
and k = 0.06, our equation becomes
$20  $0.30 e
0.06t
Let's now divide both sides by 0.30 and take the natural log of both sides.
ln
 20 
 ln e 0.06t   0.06t
0.30 
 20 
With your calculator you can find that ln 
= 4.2, so t = 70 years. With our assumptions,
0.30 
we'll be paying $20 per gallon of gas 70 years after 1960, or the year 2030 – just about the time
your children go off to college!
On the back are a few problems (with the answers given) for you to practice on.
1. Suppose you put $500 in a mutual fund with an annual growth rate of 15% (not unreasonable
these days). How long will it be before you're a millionaire? (About 51 years)
2. If inflation remains at its present rate (about 3%), how many of today's dollars will
correspond to $1 million dollars 50 years from now? (You're looking for N0. It's $223,000,
about what we used to pay the President of the US!)
3. Suppose the swing in your backyard slows down exponentially after your little brother jumps
out. If it's swinging 3 feet back and forth when he jumps out, and the size of the motion decays
at a steady rate of 5% per second, how long before it's only moving back and forth 1 inch?
(About 72 seconds)
1. N = N0 ekt with N0 = $500 and k = 0.15. We want to find t when N = $1,000,000.
1,000,000 = 500 e0.15t
1,000,000 
ln 
  ln e 0.15t   0.15t
 500 
7.6
t=
= 51 years
0.15
2. Now N = $1,000,000, k = 0.03, and t = 50 years, and we're looking for N0.
1,000,000  N0 e0,03(50)
The e0.03(50) term is just equal to 4.48, so just divide to get N0 = $223,000.
3. I’ll let you work this one. Be careful of the units. N and N0 must have the same units before
you start your mathematical manipulation to find t.