Q Skills Review
Dr. C. Stewart
Algebra 4: Exponents and Roots
Exponents
Just as we can write 3 + 3 + 3 + 3 + 3 (3 added to itself five times) as 5× 3 , so we can write
3 × 3 × 3 × 3 × 3 (3 multiplied by itself five times) using the short-hand 35 . Notice that the number
which is being multiplied (sometimes called the ‘base’) is written as usual, and the number
indicating how many of them are multiplied together is written as a superscript. We can read this
as ‘3 to the power 5’.
So what does 4 3 mean? First we recognise that this is a short-hand for multiplication, where the
only number to be written down is ‘4’. Then we look at how many times we have to write it
down – three times. So 4 3 means 4 multiplied by itself three times, i.e. 4 × 4 × 4 . 4 3 is a
representation of a number, but in order to grasp the size of this number we might want to
express its value as a base ten numeral (called a decimal number). In this case we can calculate
4 3 = 4 × 4 × 4 = 64 .
Which is larger, 3 21 or 81 6 ? To answer this we could use a calculator to find the value of each
number, or we could recognise that
81 = 9 × 9 = (3 × 3) × (3 × 3) = 34
and so 81 6 = 81× 81× 81× 81× 81× 81 = 34 × 34 × 34 × 34 × 34 × 34 .
Since each 3 4 means 3 multiplied by itself four times, then we have a total of twenty-four of
them multiplied together. Hence 81 6 = 324 .
Now we can compare 3 21 with 81 6 = 324 , and we see that 324 has more 3s multiplied together
than 3 21 . So our conclusion is that 81 6 is larger than 3 21 .
We could even go on to say that 81 6 = 324 = 33 × 321 = 27 × 321 .
We can generalise some of the reasoning used above in the following way:
If a represents any non–zero number, and m, n are any real numbers then
a m × a n = a (m + n)
and
(a )
m n
= a (m × n)
These generalisations make sense when m and n are natural numbers, since we can consider the
number of times the base, a, is multiplied by itself.
We can extend the meaning to include all integer exponents by considering that
a × a × a ( = a 3 ), divided by a, is a × a ( = a 2 ) , and so when we divide by a, we reduce the
exponent by one. This leads to the following:
2
a3 = a × a × a
a3 a × a × a
=
= a×a
a
a
a2 a × a
1
=
=a
a =
a
a
a1 a
a0 =
= =1
a
a
0
a
1
a −1 =
=
a
a
−1
1
1
a
a −2 =
= a = 2
a
a a
a2 =
1
a −2 a 2
=
= = 3
a
a a
1
a
−3
From this pattern we also see that any non-zero number raised to the power 0 is 1, and that a
negative exponent means ‘one divided by’, or ‘the reciprocal of’, the equivalent positive power.
Caution: If a = 0 then we can multiply 0 by itself any number of times and we will always get 0.
Hence a m = 0 , for any natural number m. However we cannot divide by 0, and so 0 0 , and 0
raised to a negative exponent, are undefined.
Since a − n =
1
, then we can combine this with a m × a n = a ( m + n ) to give
an
am
1
= a m × n = a m × a −n = a (m − n)
n
a
a
If m and n are integers then we can see that this makes sense. a is multiplied by itself m times in
the numerator, and n times in the denominator. We can then simplify this fraction by dividing
numerator and denominator by all common factors. If m is larger than n, we are left with a
multiplied by itself (m – n) times in the numerator.
a m a × a × a × a ×⋯ × a
=
= a (m − n)
n
a
×
a
×
⋯
×
a
a
What happens if m is smaller than n? Does the result
am
= a ( m − n ) still hold true? Here we have:
n
a
am
a × a ×⋯ × a
1
=
= ( n − m) = a −( n − m ) = a ( − n + m ) = a ( m − n)
n
a × a × a × a ×⋯× a a
a
And so we can see that the result is still true.
3
Practice Question Set 1
[Answers to these questions can be found at the end.]
First answer without using a calculator, and then try using a calculator.
1) The value of 2 4 is ______________________________
The meaning of 2 4 is ___________________________
2) Evaluate the following (remember the order of operations conventions):
2
(b) 6 2 + 4 2
(c) 3 + 2 2
(a) ( 6 + 4 )
(e) 7 32 × 7 −34
(d) 158 ÷ 157
(f)
48 × 32
46
3) Evaluate each of the following:
4
(e)  
9
(c) 2 −3
(b) 5 −1
(a) 7 0
0
1
(f)  
2
−2
(d) 4 −2
5
(g)  
3
−2
(h) 3 4
4) Complete the following by replacing ? with the appropriate value:
( )
(a) 4 × 4 × 4 × 4 × 4 × 4 × 4 = 4 ? = 22
?
5
= 2?
10
(b) 5 × 9 × 9 × 5 × 5 × 9 × 9 × 9 = ? × 5 ? = ? × 5 ?
(c) 128 ÷ 23 =
2?
= 2?
23
(d) 37 × 95 ÷ 27 2 =
( )
(3 )
37 × 3 ?
?
2
5
=
37 × 3 ?
3
?
= 3?
5) Use a calculator to find the value of the following expressions, giving your answer to 2 decimal
places.
(1 + 0.032 )( 4×5) − 1
4
20000 × ( 0.03




12 )
(a) 1500 ×
(b)
0.032
(
1 − (1 + 0.03 ) −12×4) 
( 4 )
12


So far we have looked at exponents which are integers (positive, negative or zero). Now we will
develop a meaning for an exponent which is a rational number (fraction) by considering square
roots and other roots.
4
Roots
A square root of a number is a number which when multiplied by itself gives the original
number. For example, 3 is a square root of 9, since 3 × 3 = 9 . We call 3 ‘a’ square root because
( −3) × ( −3) = 9 , and so (–3) is also a square root of 9. The term ‘principal’ is used to indicate the
positive root, and we use the symbol
(called the radical symbol) to indicate this positive
root. Hence, 9 = 3 . Note that there is no real number which when multiplied by itself gives a
negative number (in fact we say that the square root of a negative number is ‘imaginary’).
[NOTE: When solving an equation to find all square roots we write, for example, a 2 = 9 and so
a = ± 9 = ±3 . The fact that we have to write ‘ ± ’ implies that 9 indicates a positive value
only].
For a cube root, the root must be multiplied by itself three times to give the original number. For
example, the cube root of 8 is 2 because 2 × 2 × 2 = 8 . Using radical notation, we indicate the
cube root by placing a small ‘3’ beside the root symbol: 3 . Note that ( −2 ) × ( −2 ) × ( −2 ) = ( −8 ) ,
and so the cube root of (–8) is (–2). The cube root of a positive number must be positive, and the
cube root of a negative number must be negative, and so each number has only one cube root.
A fourth root must be multiplied by itself four times to give the original number, and so a
number will have both a positive and a negative fourth root. For example, the principal fourth
root of 625 is 5, since 5 × 5 × 5 × 5 = 625 , and we write 4 625 = 5 . (–5) is also a fourth root of 625,
since ( −5 ) × ( −5 ) × ( −5 ) × ( −5 ) = 625 . The fourth root has the same conditions as the square root:
a positive number will have both a positive and a negative fourth root, but there is no real fourth
root of a negative number.
A fifth root must be multiplied by itself five times to give the original number. A positive
number will have a positive fifth root and a negative number will have a negative fifth root. This
has the same conditions as the cube root.
In general we can group roots into “even” roots (square, fourth, sixth, eighth, tenth, etc.) and
“odd” roots (cube, fifth, seventh, etc.). The following table summarises the possibilities:
Even root (
Odd root (
3
4
,
,
5
6
,
,
7
, etc)
Root of a Positive Number
positive, negative
Root of a Negative Number
(not real)
positive only
negative only
, etc)
We can show the definition of roots symbolically as follows:
a× a =a
3
4
5
a×3 a×3 a =a
a×4 a×4 a×4 a =a
a×5 a×5 a×5 a×5 a =a
. . . and so on
5
1
1
Using a m × a n = a ( m + n ) , and letting m = n = 12 , we get that a 2 × a 2 = a
1
1
We have a 2 × a 2 = a , but we know that
1
1
1
Similarly, a 3 × a 3 × a 3 = a
( 12 + 12 ) = a (1) = a .
1
a × a = a , and so we can conclude that a 2 = a .
( 13 + 13 + 13 ) = a (1) = a and 3 a × 3 a × 3 a = a , so a 13 = 3 a .
1
In general we can say that a r = r a , for any natural number r.
( )
If we combine this with a m
n
ar = a
( )
( 1r × n ) = a 1r
n
=
n
= a ( m × n ) , then we can understand any rational exponent:
( )
r
a
n
n
and also
ar = a
( n × 1r ) = a n 1r = r a n
( ) ( )
A Special Note: It is important to know how to find values involving exponents using a
calculator. For different calculators there are differences in the order in which you must type the
numbers and in the symbol on the buttons. Most calculators have a special button x 2 , but this
can be used only for ‘squaring’. Look for x y , or y x , or simply ∧ to enter other exponents. For
‘square root’ there is usually a button marked x . For other roots you can use the exponent
button and enter a fraction as the exponent (you may need to use brackets around the fraction), or
your calculator may have one of the following buttons:
important to become familiar with your own calculator.
y
x , or
x
1
1
y , or x y , or y x . It is
Working with Roots
The square root of a number may be a natural number, for example
e.g.
9 = 3 , or a rational number,
2.25 = 1.5 = 3 . But the square root of some numbers is irrational, for example,
2 . In fact
2
the square root of any prime number is irrational. For irrational square roots we can either give a
decimal approximation to an appropriate precision (e.g. 2 ≈ 1.414 ) or leave the
symbol to
keep the exact value.
When roots are multiplied or divided it can be useful to combine or separate them in order to
simplify expressions.
a
a
Combining roots we have:
a × b = a × b and
=
b
b
a×b = a × b
Separating roots we have:
and
a
a
=
b
b
How can we be sure this is correct? By definition, if we multiply a square root by itself then we
get the original value. Now
(
) (
a× b ×
)
a × b = a × a × b × b = a × b . So
(
a× b
)
6
multiplied by itself gives a × b , which means that “the square root of a × b is
(
)
a × b ”, which
in symbols is written a × b = a × b . Since the two expressions are equal we can also write it
the other way round:
a × b = a × b . We can use the same type of argument to prove the
relationships for division of roots.
An example of using this ability to separate roots is to simplify the square root of 45:
45 = 9 × 5 = 9 × 5 = 3 × 5 = 3 5 .
We can check this using the definition of the square root of 45 (a number which when multiplied
by itself gives 45): 3 5 × 3 5 = 3 × 5 × 3 × 5 = 3 × 3 × 5 × 5 = 9 × 5 = 45 .
An example where combining roots allows us to simplify is as follows:
44
27
44 27
×
=
×
15
55
15 55
It is important to pause at this point to consider what to do next. It would be hard work to
multiply the numerators and denominators! In general, in order to simplify we want to look for
factors and thus work with smaller numbers rather than creating very large numbers. In this
example we can simplify (cancel) the fractions to arrive at a much simpler root:
44 27
4 × 11 3 × 9
4×9
×
=
×
=
5×5
15 55
5 × 3 5 × 11
Finally, we can separate the roots and evaluate them:
4×9
4×9
4 × 9 2×3 6
=
=
=
=
5×5
5
5
5
5× 5
44
27
6
×
would simplify to ?
15
55
5
Who would have thought that
Practice Questions Set 2
[Answers to these and questions raised above can be found at the end.]
1) Change between radical notation
(a) x
(e) r
(b)
1
2
(f) t
4
−
m
1
3
( a ) and exponent notation ( a
(c)
r
3
1
p
(g) 6 x
1
r
):
(d) 2 3 g
1
1
4
(h) ( 9x ) 2
2) Evaluate each of the following:
1
1
(a) 27 3
(b) 25 2
(g) 10
−3
(h) 64
1
6
(d) 3− 2
(c) (0.4) −1
 25 
(i)  
 49 
−
1
2
−
(j) 81
(e) (0.008)
1
2
−
1
3
(f) 16
−
1
1
(k) (0.125) 3
(l) 32 5
1
4
7
3) Evaluate the following:
(a) 12
0
1
− 42
3
3
4) Simplify the following:
(a) 8
(b) 180
49
36
(e)
8
50
(f)
3
(c) 16 2 + 16 4
(b) 25 2
(c)
84
(d)
14 × 42
(g)
12 × 15
30
(h)
10
35
×
21
39
Answers
Practice Questions Set 1
1) The value of 2 4 is 16. The meaning of 2 4 is 2 × 2 × 2 × 2
2) (a) ( 6 + 4 ) = (10 ) = 100 (b) 6 2 + 4 2 = 36 + 16 = 52 (c) 3 + 2 2 = 3 + 4 = 7
2
2
(d) 158 ÷ 157 = 15(
(f)
8− 7 )
(e) 732 × 7 −34 = 7(
= 151 = 15
32 + ( −34 ) )
= 7(
32 −34 )
1
1
−2
= 7( ) = 2 =
49
7
48 × 32
8−6
= 4( ) × 32 = 4 2 × 32 = 16 × 9 = 144
6
4
0
1
(f)  
2
4
(e)   = 1
9
(h) 34 = 3(
(c) 2 −3 =
(b) 5 −1 = 15 = 0.2
3) (a) 7 0 = 1
2+ 2)
−2
=
1
( )
1 2
2
=
1
( )
1
4
1
23
5
(g)  
3
= 1× ( 14 ) = 4
= 32 × 32 = 9 × 9 = 81 or 34 = 3(
2×2 )
(d) 4 −2 =
= 81 = 0.125
( )
= 32
2
−2
=
1
( )
5 2
3
=
1
42
1
= 161 = 0.0625
( )
25
9
=
= ( 9 ) = 81
2
4) (a) 4 × 4 × 4 × 4 × 4 × 4 × 4 = 47 = ( 22 ) = 214
7
(b) 5 × 9 × 9 × 5 × 5 × 9 × 9 × 9 = 95 × 53 = 310 × 53
27
(c) 128 ÷ 23 = 3 = 24
2
(d) 3 × 9 ÷ 27 =
7
5) (a)
(b)
5
2
37 × ( 32 )
(3 )
3 2
5
=
37 × 310
= 311
6
3
(1 + 0.008 )( 20 ) − 1

 ≈ 1500 × [ 0.172764 ] = 32393.26 to 2 d.p.
1500 × 
( 0.008)
( 0.008 )
20000 × ( 0.0025 )
1 − (1.0025 )

( −48 ) 

≈
50
50
≈
≈ 442.68654 = 442.69 to 2 d.p.
[1 − 0.8870533] 0.1129467
9
25
= 0.36
8
Practice Questions Set 2
1
x = x2
1) (a)
(f) t
2) (a)
−
1
3
(b)
= 27 = 3
(e) (0.008)
(c)
1
3
(b)
 8 
=

 1000 
−
1
25 2
1
3
1
−1
=p 3
3 p
1
1
(d) 2 3 g = 2 g 3
(e) r 2 = r
1
(h) ( 9 x ) 2 = 9 x = 3 x
(g) 6 x = 6 4 x
3
−
1
m = m4
1
4
1
=3
t
1
27 3
4
(c) (0.4) −1 =
= 25 = 5
1 10
=
= 2.5
0.4 4
1
 1000  3 3 1000 10
=
=
=5
 = 3
2
8
 8 
(f) 16
−
1
4
(d) 3− 2 =
=
4
1
32
= 19
1
1
= = 0.5
16 2
1
(g) 10 − 3 =
 25 
(i)  
 49 
−
1
1
=
= 0.001
3
10 1000
1
2
 49 
= 
 25 
1
(k) (0.125) 3 =
3) (a) 12
(c)
0
3
16 2
1
− 42
3
1
2
=
(h)
−
49 7
= 5 = 1.4
25
(j) 81
3
3
 1  1
= 16 2  + 16 4  =

 

=
1
2
1 1
×
64 2 3
=
1
1
81 2
1
 1 3
=  64 2  = ( 8 ) 3 = 2




=
1
9
1
125
5 1
=
= = 0.5
1000 10 2
(l) 32 5 = 5 32 = 5 4 × 8 = 5 2 × 2 × 2 × 2 × 2 = 2
= 1 − 4 = 1 − 2 = ( −1)
3
+ 16 4
1
64 6
(b)
3
25 2
( 16 ) + ( 16 )
3
4
3
3
 1
=  25 2  =


(
25
)
3
= ( 5 ) = 125
3
= ( 4 ) + ( 2 ) = 64 + 8 = 72
3
3
8 = 4× 2 = 4 × 2 = 2 2
4) (a)
(b)
180 = 9 × 20 = 9 × 4 × 5 = 9 × 4 × 5 = 3 × 2 × 5 = 6 5
(c)
84 = 4 × 21 = 4 × 3 × 7 = 2 × 3 × 7 = 2 21 * we consider the square root to have been
simplified if the number under the
has no factor which is the square of a rational number.
(d)
14 × 42 = 14 × 42 = 2 × 7 × 2 × 3 × 7 = 2 × 2 × 3 × 7 × 7 = 2 × 7 × 3 = 14 3
(e)
49
49 7
=
=
36
36 6
(g) =
(h)
(f)
8
=
50
4× 2
=
25 × 2
4× 2
25 × 2
=
2
= 0.4
5
12 × 15
12 × 15
12
=
=
= 6
2
30 2
30
10
35
10 35
2× 5× 5× 7
5× 5 2× 7
5 2
×
=
×
=
=
×
=
21
39
21 39
3 × 7 × 3 × 13
3 × 3 7 × 13 3 13