Inactive Tutor answered • 06/19/15
Tutor
New to Wyzant
We need to use the exponential growth formula:
P(t) = Poekt
P(t) is the amount over time = $455
Po is the starting amount = $231
k is our rate of growth (to be found)
t = time = 12 years (1987-1975 = 12 years)
455 = 231e12k
e12k = 455/231
Take the natural log of both sides
ln(e12k) = ln(455/231)
12k ln(e) = ln(455/231)
Note that ln(e) = 1 so we have
12k = ln(455/231)
k = ln(455/231)/12
k = .0565 or 5.65%
Our exponential growth rate is approximately 5.65%
Inactive Tutor
I've been looking through my old answers and, although this one is several months old, I feel I ought to explain a bit about exponential growth just in case another student happens upon it.
First, it's important to note that there's no specific "exponential growth rate formula." Exponential growth describes any situation in which something is increasing exponentially. That is, exponential growth occurs whenever you see the variable in the exponent. So, a basic exponential growth formula looks like this: P(t) = a(r)t. Replacing r with the constant e gives you one exponential growth formula, usually used for computing continuously-compounded interest. It's not really applicable in this particular case.
Second, in any exponential growth formula, you can immediately pick out the rate of growth r and the initial amount a. So, P(t) = 50(1.2)t models a situation where we started with 50 of something and it grew at a rate of 1.2, or 120%. That is, it increased 20% each year. You can tell when a quantity is increasing because the rate will be over 1. Anything under 1 indicates an exponential decrease.
Since the rate always appears in a specific place in the formula (in the base, not in the exponent), you should not solve for a variable in the exponent and call it the rate, as Andrew does. And, if the rate of growth for something you know is increasing comes out to be less than one (as in Andrew's answer, where k=0.0565), you ought to check your work---this is a good indication that you've done something wrong.
Therefore, please use the second answer (below) as a guide for problems like this, not the first.
04/23/16