College Algebra
Answer/Discussion to Practice Problems
Tutorial 47: Exponential Growth and Decay
Answer/Discussion
to 1a
The value of the property in a particular
block follows a pattern of exponential growth. In the year 2001,
your company purchased a piece of property in this block. The
value of the property in thousands of dollars, t years after 2001 is
given by the exponential growth model 
A)
What did your company pay for the property? 
Since we are looking for the value of the property,
what variable are we seeking? If you said V you are right
on!!!!
Note that even though we are using V instead of A, the
concept of and formula for exponential growth is the same as discussed
in the lesson.
The way the problem is worded, 2001 is what we call our initial
year. This is where t = 0.
Plugging in 0 for t and solving for V we get: 

*Replace t with 0
*e^0 simplifies to be 1

While writing up the final answer, keep in mind that
the problem said the value was in thousands of dollars.
In 2001, the value of the property was $500,000.
We could have also approached this problem by noting that the year was
2001, which is our initial year, so basically it was asking us for the
initial worth, which is Vo (Ao in the formula).
This happens to be the number in front of e which is 500 in
this problem.
The reason I showed you using the formula was to get you use to
it. Just note that when it is the initial year, t is 0,
so you will have e raised to the 0 power which means it will
simplify to be 1 and you are left with whatever Vo (Ao)
is. 
B) By what percentage is the price of the property
in this block
increasing per year? 
In the general growth formula, k is a constant
that represents the growth rate. k is the coefficient of t in e’s exponent.
So what would be our answer in terms of percent?
Well, k = .055, so converting that to percent we get 5.5% for our answer.

C) What will the property be worth in the year 2010? 
Since we are looking for the population, what variable
are we looking for? If you said V give yourself a high
five.
What are we going to plug in for t in this problem?
Our initial year is 2001, and since t represents years after
2010, we can get t from 2010  2001, which would be 9.
Plugging in 9 for t and solving for V we get:


*Replace t with 9
*e^ .495 is
approx. 1.640498239

Keep in mind that the problem said that the value was
in thousands of dollars.
The value of the property in 2010 would be approximately
$820,249.12. 
D) When will the property be worth 750 thousand
dollars? 
Again, it looks like we have a little twist here.
Now we are given the value of the property and we need to first find t to find out how many years after 2001 we are talking about and then
convert that knowledge into the actual year.
We will still be using the same formula we did to answer the questions
above, we will just be using it to find a different variable.
Plugging in 750 for V and solving for t we get: 
This means a little over 7 years after 2001, the value
of the property will be 750 thousand dollars.
So our answer is during the year 2008. 
Answer/Discussion
to 1b
An artifact originally had 10 grams of carbon14
present. The decay model describes the
amount of carbon14 present after t years.

A) How many grams of carbon14 will be present in
this
artifact after
25,000 years? 
What are we going to plug in for t in this
problem?
Since t represents the number of years, it looks like we will
be plugging in 25,000 for t.
Plugging in 25000 for t and solving for A we get:

There will be approximately .49 grams of carbon14
present after 25,000 years. 
B) What is the halflife of carbon14? 
If we are looking for the halflife of carbon14, what
variable are we looking for? If you said t give yourself
a high five!!!!
It looks like we don’t have any values to plug into A.
However, the problem did say that we were interested in the HALFlife,
which would mean ½ of the original amount (10) would be
present at the end (A). This means A can be
replaced with 10/2 = 5.
Replacing A with 5 and solving for t we get: 
The halflife of carbon14 is approximately 5728
years. 
Last revised on March 23, 2011 by Kim Seward.
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