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 = 1/143 = 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) = Ao(1 + r)t
Where Ao 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:    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 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
                                         Ao = 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 44.  This greatly simplifies the problem!!
That's about it for this section!!
 
Let's head on to the next section!