Transcript Exponent Laws

Exponent Laws
Terms to know before starting this topic
Power: an expression containing a number or variable,
which is called the base, with an exponent. For
example:
2
x
Exponential Expression: an expression containing
any power or powers. An example is:
x 2 y 3  x3 y 5
Simplified Exponential Expression
• Any expression in mathematics usually has to be simplified.
Exponential expressions are no exception. To simplify any expression,
the conditions of a simplified expression have to be known. For an
exponential expression, there are four conditions that need to be met to
have a simplified answer. These four conditions are:
• No base can appear more than once.
• No base can have a negative exponent
• No base can have a zero exponent
• Answer should be a single rational
An exponential expression is simplified with the help of the exponent laws.
There are seven of these laws. Each law specifically fixes an expression to
satisfy one of the four conditions of a simplified exponential expression.
For instance, the first condition requires that a simplified answer has
only one of any base. Exponent laws 1 and 2 are used to make two or
more of the same base change to a single base.
Below is an expression that has more than one base of x:
x 2  x3
This expression can be simplified by rewriting the expression in
expanded form and then just multiplying.
x  x  x  x    x  x  x   x
2
3
5
• So, here’s the simplified result:
x x  x
2
3
5
So the result has a single base which is the same
base as the original two powers but has a new
exponent of five. The new exponent is just the
sum of the original two exponents. The first
exponent law just describes this idea of adding
exponents.
Exponent Law #1
x x  x
m
n
m n
Thus, when using Law #1, the BASE DOES NOT CHANGE!
Only the exponents of the bases are added. Below are some
examples of using this law:
3
7
37
10
y

y

y

y
1.
4
2
4 2
6
2

2

2

2
2.
Note in the second example that the base of 2 stayed in the
answer. You DO NOT multiply the twos to get a four.
2
8
1 1 1
3.        
 2  2  2
2 8
10
1
 
 2
Note that the brackets do not disappear in the final
answer!
3
3
33
6









4


4


4


4
4.
Note: keep the brackets around the negative number!
5.
x 3  x 4  x 5  x 3 45  x12
Note: the law works for three or more of the same base.
Just add all exponents of the like base.
6. If there are two or more bases, add the exponents of the bases
that are the same, as shown below.
x y x y  x
3
2
4
6
3 4
y
2 6
x y
7
8
7. Be careful of a case where there is no exponent. When
there is no exponent shown, the exponent is a one.
a a  a a  a
7
7
1
7 1
a
8
8. If there is not more than one of a base, leave the base
alone. An example of this is:
x y x  x
2
3
3
2 3
y x y
3
5
3
Let’s look at a different case. What if the operation
between two bases is division.
For instance, simplify the following expression:
x x
7
3
Since all divisions can be written as fractions, this
expression could be changed to:
7
x
3
x
Now remember, the expression is NOT simplified since
there are two of the same base. We need to simplify the
expression. First, change to expanded form:
x
x x x x x x x

3
x
x x x
7
Now, we can cancel the x’s that are on top with the x’s on
bottom to leave just four x’s on top, like this:
x
x x x x x x x
4


x

x

x

x

x
x3
x x x
7
So, this means that
x x  x
7
3
4
Note that all we really did was subtract the 3
from the 7. This will always work when
dividing the same base. Thus we could have
simplified by the following method:
x x  x
7
3
73
x
4
So, now we have a new exponent law!
Exponent Law #2
m
x
mn
x x  n  x
x
m
n
Let’s use this new law to simplify expressions.
y11
11 7
4

y

y
1.
y7
75
5 3
2
2.

7

7
73
Again, as in law one, the base does NOT
change. The 7 stays and does not get
cancelled. Also, the answer can be changed
to 49, but this is not required.
What if there was two different bases. Then just subtract the
exponents of the like bases, as shown below.
x6 y8
62
8 3
4
5

x
y

x
y
3.
x2 y3
4.
a 9b 3
a 9b 3
 7 1  a 97b 31  a 2b 2
7
a b
a b
Don’t forget that if a base does not have an
exponent, the exponent is really one, NOT ZERO!
5.
p 4q3
 p 43 q 3  p1q 3  pq3
3
p
Remember that if only one of a base appears, do nothing
to that base. Since there is only a q in the numerator,
leave it alone!
Before doing Law #3, let’s recall using powers from
earlier grades.
What if we had to simplify:
4 4
5
5
Like any division, this can be changed to
fractional form. Thus, the expression could
appear as the following:
5
4
5
4
Let’s simplify using expanded notation:
4
44444

5
4
44444
5
Since there are fours in the numerator and
denominator, the fours just cancel each other
out to make a bunch of ones, as so:
11111 1
 1
11111 1
We could also simplify the expression using Law
#2 that says to just subtract the exponents. Thus,
the question could be done using the following
approach:
45
5 5
0
4 4
5
4
Since the expression was done correctly in
both methods, then their results are
equivalent. So,
4 1
0
So, now we have a new law!
Exponent Law #3 - The Zero Exponent Law
x 1
0
So, Law #3 states that a power of ANY base with an exponent of zero, can
be replaced with a one. This law is then used for eliminating bases with
exponents of zero that cannot occur in a simplified exponential expression
as mentioned on slide 2.
Here are some examples of using this law:
1.
x0  y 0  1  1  2
2.
10000  1
3.
0
x0 1

10 10
Before moving on to Law #4, let’s go back to Law #2 for a
second. All the examples using this law had the numerator
base with a higher exponent than the denominator base. What
if this wasn’t the case? Let’s see!
Take the case below;
3
x
5
x
This expression can be solved two ways.
Let’s first do the question by using expanded forms
and canceling bases.
x3
x x x
111
1


 2
5
x
x  x  x  x  x 111 x  x x
Now, let’s do the question using the rule from
Law #2.
3
x
3 5
2
x x
5
x
Since both methods are correctly done, then the
only conclusion that we can make is that the
two results are equal. Thus;
x
2
1
 2
x
As was stated earlier, an expression cannot
contain a negative exponent, so we now have a
way of eliminating negative exponents by
reciprocating the base. In the case above, the x
with the negative exponent was reciprocated and
the x is in the denominator but with the exponent
changed to its positive opposite. We now have a
new law!
Exponent Law #4
The Negative Exponent Law
x
m
1 1
n
 m , n  x
x x
It is important that the base does NOT CHANGE when
using this law. The base really just flips. Here are some
examples of using this law:
3
3
1
1
y
y
 2
1. x 2 y 3  2 y 3  2 
x
x
1 x
This question can be shortened by just realizing that
the base with a negative exponent just moves from
the numerator to denominator or vice versa. Let’s do
the next example this way!
2.
x 3 y 1
z4
z4
 3 1 or 3
4
z
x y
x y
So, as you can see, each of the three bases just
switched from top to bottom or reverse.
3.
x 2
z5
 2 3
3 5
y z
x y
Notice that the y base did not move. This is because the y does not
have a negative exponent and thus, does not need to be changed.
4.  3 x
6
 3 x
1
6
3
 6
x
Note that the -3 did not move. This is because a negative number is
acceptable but a negative exponent is not, thus x moved to the
denominator. The -3 has an exponent of one, not a negative exponent.
1
1
1
3



3

or


5.
 33  27 27
Again, as was stated several slides ago, don’t remove the brackets until
you are evaluating. The -3 kept the bracket as the number moved to the
denominator.
Before moving on to Law #5, let’s again do some examples
that are simplified by expanded notation.
x 
3 2
 x3  x3  x  x  x  x  x  x  x 6
Since squaring means to multiply a term by itself, we
3
multiplied the x by itself. Then expand to get six x’s
to find our result.
Also, note that the resulting exponent is just the product
of the two exponents in the original expression. This
concept will always work when the is a “power of a
power” with brackets. So now we have a new law!
Exponent Law #5
The Power of Power Rule
x 
m n
x
m n
x
mn
So let’s use this new law to simplify some expressions.
1.
3 
3 4
34
3
3
12
Note that the base of 3 does NOT CHANGE! Again, rarely
does a base change unless you’re evaluating. In this case, the
answer is a large number, so don’t evaluate!
2.
a 
3.
x 
3 2
2 4
a
x
32
2 4
a
x
8
6
1
 8
x
In case #3, after multiplying the exponents, the new
exponent was still negative which meant that the base has
to move to the denominator using Law #4. Thus, this
question used two laws to simplify completely.
All Law #5 is used for is to eliminate brackets. But what if
there is more than one base inside the bracket? Let’s see!
x y 
4 2
3
 2x
 3
 y
2
 x y x y  x y
3
4
3
4
6
8
3
 21 x 2 21 x 2 21 x 2 23 x 6
  3  3  3  9
y
y
y
y

So, as you can see, each inside exponent is just multiplied
by the outside exponent to get the final exponent. Notice
that none of the bases changed!
So, now we have two new exponent laws to help
eliminate brackets!
Exponent Law #6
x y 
m
n a
x
ma
y
n a
x y
ma
Exponent Law #7
x
 n
y
m
a

x
x
  na  na
y
y

m a
ma
na
Let’s do some examples using laws 5 - 7.
1.
x y 
2.
3x   3 x 
2
5 3
 x 23 y 53  x 6 y15
3 2
1
3 2
12
3 x
32
 3 x  9x
2
6
6
Remember that any base without an exponent, such as the 3
above, that the base actually has an exponent of one, not zero!
3.
4
3 4
12
 x 
x
x
 2   24 14  6 4
y z
y z
 y z
3
Here are some examples using more than one law in a question.
1.
x
2
y

3 5
15
y
 x 10 y15  10
x
44  46
44 6 42
1
1
2. 2 3  2( 3)  5  2 5  7
4 4
4
4
4 4
4


1
3. x  y  z  3  1  1  1  3  3  3  3  3
3
0
0
0
4
4
4
3
2
 x y 
x 6 y 2
1
1
4.   2  4   4 8  4( 6) 82  10 6
x y
x
y
x y
x y 
3
5. 4 x

3
1

y  2 2 xy 4
4
1

8 y14
2 y14

3
 41 x 3 y  2  2 3 x 3 y 12  41 x 0 y 14  2 3 
4
23 y14
Here are some special examples!
1.
3  6  3  3  6  6  324
2
2
Since the bases are different, the exponent laws CANNOT be used. Thus,
we refer back to expanded form and then just multiply together.
1
1
2
x 
y 
y
2.  2    3   3
y 
x 
x
 
 
3
2
If a fraction has a negative exponent, just reciprocate the fraction to change the
sign of the exponent. Then get rid of brackets as in Law #7.
1
1
2
3
1 1 3 2 5
3. 2  3 

 1 1  
1
1
2 3 6 6 6
1
1
Since there is an addition sign, the bases must be treated as separate fractions,
not one fraction. Then add the fractions using a common denominator.
Try the following questions. Answers are on the next slide.
Simplify the following expressions
completely.
4
x x y y x
2
3
2
x   x 
3 3
1  2
a b 
a b 
Evaluate the following:
50  61
7 2
2 1  31
4 1
3 2 2
 32   35
3 2
3x y  6 xy 
2
2
5
5 5
5  510
3
 3 1
x 0  2 y 0  31
2
 
5
4
5
 
2
3
Here are the answers!
6
x
y2
1
x11
a 5b 2
3y5
2x5
1
53
7 3  343
2
3
1
27
2
3
5
2