 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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# Exponential Growth and Decay - Problem 1

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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I want to talk about exponential growth again. This time I’m going to give you the exponential growth formula, n of t equals n sub 0 times b to the t. This is an exponential function with a constant in front, this n sub 0 represents some initial amount. It could be an initial population, an initial amount of money, anything like that.

Here is the problem, the tuition at Hogwartz has doubled every century, what is the annual rate of increase? If the tuition is now 2400 galleons per year, what would it be in 10 years? So here we are starting with the doubling time, that tells us how long it takes for the tuition to double, 100 years. So let me write my formula.

Now here I could use the initial amount of 2400 galleons, but it doesn’t actually matter. So I’m going to plug in, 2 times n sub 0 equals n sub 0 b to the t. Now the doubling time is actually a century, 100 years. So this t value should be 100.

I need to solve for b, the base. So I divide both sides by n sub 0. Remember I already know that the initial number is going to be \$2400. Dividing that out gives me 2 equals b to the 100. I need to solve an equation like this. You take the 100 through of both sides.

Of course that’s the same thing as raising to the 1 over 100th power. So I’m going to do that on my calculator, I’ll write that down though. 2 to the 1 over 100, that’s your b value. So 2 to the 1 over 100, b is approximately 1.00695555.

When I do any calculations using this value, I’m going to want to use the stored value. So I’m going to keep that stored. The annual rate of increase comes from this number, this is the multiplier I have to multiply the tuition by this number every year. The 1 gives me the original tuition, and this part gives me the increase. So the increase, is .0695555 times the initial amount. In percent, this would be .695555 percent. So not quite .7 percent per year. That’s the rate of increase.

Now I want to find out, what will the tuition be in 10 years. Well here is my model, n sub 0 is 2400 galleons. The b value, the base, is 1.00695555. And I want to find out what’s going to happen in 10 years. So for the t, I plug in 10 and then it’s just a calculation. 2400 times, I stored this value to the 10th power. This gives me 2572.26. 2500, 72 galleons and sub circles.