Section 5-1:
Integral
Exponents
*Basic Laws of
Exponents *
*Each of the above rules
should
be familiar to you from algebra I.*

*Here are some sample
problems
with their solutions.*

*1) Watch the
difference
between these two:*

*
a) (-3)*^{-2 }
b) -(3)^{-2}

*
The first one is squaring a negative number and the second is squaring
a positive number and then making the whole result negative.*

*
a) = 1/(-3)*^{2} =
1/9
b) 1/-(3)^{2} = -1/9

*
c) 7*^{ .} 2^{-3}
=
d) (7^{ .} 2)^{-3} =

*
The first one raises the power then multiplies, while the second one
multiplies
first then raises the power.*

*
c) =
7/8
d) = 14*^{-3} = 1/14^{3} = 1/2744

*In the above
example
our first step is to work inside the grouping symbols and get a common
denominator. Then add the two fractions. Only when you have
a single fraction, is it permitted to invert the fraction.*

*In the above
example
we again simplify inside the grouping symbols and get a common
denominator.
Once we have a single fraction in step 3 we can invert the
fraction.
Notice the factoring in the last step!*

*In this example,
the
numerator and denominator have the same base. We can apply the
division
rule by subtracting the exponents. Then simplify. Remember,
no negative exponents should be left in the answer.*

*An important type
of
rule can now be stated using exponents. It is a growth or decay
problem.
We can mathematically model this function by using the following:*
*A(t) = A*_{o}(1
+ r)^{t}
*Where A*_{o}
is
the initial amount at time t = 0
*r is the rate (as a
decimal)*
*t is the time*
*A(t) is the amount
after
the time t.*
*If r > 0, then
it is
an exponential growth.*
*If -1 < r <
0,
then it decays exponentially.*

*1) Suppose a
bike
costs $100 now and it increases at a rate of 5% per year. What
will
be the cost of the bike in 4 years?*
* *
*
Solution: A*_{o} = 100, r = 5% = .05, t = 4

*
A(4) = 100(1 + .05)*^{4 }= 100(1.05)^{4} = 121.55

*The bike will cost $121.55
in 4 years. (Rounded to the nearest penny)*

*2) Suppose a
car
is worth $15,000 new. What will it be worth in 3 years if it
decreases
at a rate of 12% per year?*
* *
*
Solution: This is a decrease problem with*

*
A*_{o} = 15000, r = -.12, t = 3

*
A(3) = 15000(1-.12)*^{3} = 15000(.88)^{3} = 10222.08

*The car will be worth
$10222.08
in 3 years.*

One of the easier ways to
do
this problem is to multiply both numerator and denominator by the
positive
power of the biggest negative exponent. In this case we
multiplied
by 4^{4}. This greatly simplifies the problem!!

*That's about it for
this
section!!*
* *
*Let's head on to
the
next section! *
* *