# FACT: The formula can be used for every exponential growth and by EhZMOG

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```									 Math 90 – Exponential Growth and Decay Problems and Formulas

FACT: The formula A  Pe rt can be used for every exponential growth and decay problem you
face, including money, radioactive decay, population growth, doubling-time, half-life, tripling-
time, etc. In fact, if a word problem doesn’t fit one of the models below, you’ll need to revert
back to good ‘ol “Pert.”

BUT…

Check this out:

t/h
1
For decay problems, try: A(t )  A0            where h = the half-life of the element
 2

For “doubling time” problems, try: A(t )  A0  2t / d where d = time to double

For “tripling time” problems, try: A(t )  A0  3t / T where T = time to triple

Get the idea? Do you see where the “half,” “doubling,” and “tripling” words play out?
Do you see how the exponent works out?

Example 1: Half-life of Sandersium is 3 weeks, so if you inject some Sandersium into your
system, in three weeks only half of it will remain. So the model is:

t /3
1
A(t )  A0  
 2

3/ 3         1
1           1    1
And then in three weeks, you’ll have A(3)  A0            A0    A0
 2           2   2
in your head. That is, half of the initial amount is in there in two weeks, exactly as
expected.

Example 2: The doubling time of dorkism is 4 weeks. That is, in a dorky environment with a
given starting amount of dorkism in your system, that will double every 4 weeks, so you’ll be
twice as dorky as you were four weeks ago. So the model will be:

A(t )  A0  2t / 4
So in four weeks, you’ll have A(4)  A0  24 / 4  A0  21  2 A0 . As expected, in four
weeks you’ll have twice your initial dorkiness.

So, how much dorkiness will you have at the end of a 16-week semester? That’s four
weeks, four times, right? So you’d expect to double four times… and 24  16 . So you
should be 16 times as dorky as when you started. Check the math:

A(16)  A0  216 / 4  A0  24  16 A0

So, if a student drops out of Sanders’ class after only 5.5 weeks, how much dorkier is that
student than when (s)he started the course?

Answer: A(5.5)  A0  25.5 / 4  A0  21.375  2.594 A0 . So the lucky student is only about
2.6 times as dorky as when (s)he started the course.

So what about those poor souls who take both Math 70 and Math 90 with Sanders and
get exposed to 32 weeks of dorkiness?

Answer: A(32)  A0  232 / 4  A0  28  256A0 - They’re 256 times as dorky as before
they started! Scary, eh? But that explains some things too, doesn’t it?! 

Example 3: A certain mobster loans out money, but expects triple payback – with a tripling time
of 5 weeks. If you take out a loan with this monster, and don’t pay it back for 5 weeks, you’ll
owe him triple. If you don’t pay back for 10 weeks, you pay back triple of triple – or 9 times the
original amount. So the model is:

A(t )  A0  3t / 5

And again you can see that plugging in t = 5 gives:

A(5)  A0  35 / 5  A0  31  3 A0

So if you borrowed \$250 from this mobster monster at the beginning of the 32-week
math 70-math 90 sequence to buy your math book and pay for the units for both courses
while you were continually being infected with dorkiness and Sandersium was slowly
decaying in your system, how much would you owe the guy at the end of Math 90?

Answer: A(32)  250  332 / 5  250  36.4  250 1131.295  282823.86

That is, you’d owe the guy \$282,823.86!! Now that’s a serious chunk of change!

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