Basic Laws of
Each of the above rules
be familiar to you from algebra I.
Here are some sample
with their solutions.
1) Watch the
between these two:
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 =
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
first then raises the power.
d) = 14-3 = 1/143 = 1/2744
In the above
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
we again simplify inside the grouping symbols and get a common
Once we have a single fraction in step 3 we can invert the
Notice the factoring in the last step!
In this example,
numerator and denominator have the same base. We can apply the
rule by subtracting the exponents. Then simplify. Remember,
no negative exponents should be left in the answer.
An important type
rule can now be stated using exponents. It is a growth or decay
We can mathematically model this function by using the following:
A(t) = Ao(1
the initial amount at time t = 0
r is the rate (as a
t is the time
A(t) is the amount
the time t.
If r > 0, then
an exponential growth.
If -1 < r <
then it decays exponentially.
1) Suppose a
costs $100 now and it increases at a rate of 5% per year. What
be the cost of the bike in 4 years?
Solution: Ao = 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
is worth $15,000 new. What will it be worth in 3 years if it
at a rate of 12% per year?
Solution: This is a decrease problem with
Ao = 15000, r = -.12, t = 3
A(3) = 15000(1-.12)3 = 15000(.88)3 = 10222.08
The car will be worth
in 3 years.
One of the easier ways to
this problem is to multiply both numerator and denominator by the
power of the biggest negative exponent. In this case we
by 44. This greatly simplifies the problem!!
That's about it for
Let's head on to