1. Introduction
If a is any number and n any positive integer (whole number) then the product of a with itself n times, a _{1} × a _{2} × ... × a_{n} , is called a raised to the power n , and written a^{n} ,
i.e.:
a^{n} = a _{1} × a _{2} × ... × a_{n}
Example 1:
(a) 7^{2} = 7 × 7 = 49
(b) 2^{3} = 2 × 2 × 2 = 8
(c) 3^{5} = 3 × 3 × 3 × 3 × 3 = 243
The following important rules apply to powers:
Rule 1
a^{m} × a^{n}
=
a^{m + n}
Rule 2
a^{m} ÷ a^{n}
=
a^{m − n}
Rule 3
(a^{m} )^{n}
=
a^{m × n}
Rule 4
a ^{1}
=
a
Rule 5
a ^{0}
=
1
We want these rules to be true for all positive values of a and all values of m and n . We shall first look at the simpler cases:
Example 2:
(a) 10^{2} × 10^{3} = (10 × 10) × (10 × 10 × 10) = 10^{5} = 10^{2+3}
(b) 2^{5} ÷ 2^{3} = 32 ÷ 4 = 2^{2} = 2^{5−3}
(c) (3^{2} )^{3} = (3 × 3)^{3} = (3 × 3) × (3 × 3) × (3 × 3)
= 3^{6} = 3^{2×3}
(d) From Rule 2, a ^{n+1} ÷ a^{n} = a ^{1} . Also:
a ^{n+1} ÷ a^{n} = a _{1} × a _{2} × a _{3} × ... × a _{n+1} / a _{1} × a _{2} × a _{3} × ... × a _{n} = a = a ^{1}
(e) We have a^{n} × a ^{0} = a ^{n+0} = a^{n} = a^{n} ×1 . Thus a ^{0} = 1 .
Exercise 1: Simplify each of the following.
(a) 2^{3} × 2^{3}
Solution: 2^{3} × 2^{3} = 2^{(3+3)} = 2^{6} = 64
(b) 3^{15} ÷ 3^{12}
Solution: 3^{15} ÷ 3^{12} = 3^{(15−12)} = 3^{3} = 27
(c) (10^{2} )^{3}
Solution: (10^{2} )^{3} = 10^{(2×3)} = 10^{6} = 1,000,000
Click on questions to reveal their solutions
2. Negative Powers
The question now arises as to what we mean by a negative power. To interpret this, note that
a ^{2} ÷ a ^{5} = a ^{2} / a ^{5} = a × a / a × a × a × a × a = 1 / a ^{3}
If Rule 2 is to apply, then a ^{2} ÷a ^{5} = a ^{2 − 5} = a ^{−3} . Thus a ^{−3} = 1 ⁄ a ^{3}
The general rule is:
Example 3:
(a) 10^{−2} = ^{1} ⁄_{102} = ^{1} ⁄_{100}
(b) 3^{−1} = ^{1} ⁄_{31} = ^{1} ⁄_{3}
(c) 5^{2} ÷ 5^{4} = 5^{2 − 4} = 5^{−2} = 1 ⁄ 5^{2} = ^{1} ⁄_{25}
Exercise 2: Write each of the following in the form a^{k} , for some number k .
(a) 2^{3} × 2^{−5}
Solution: 2^{3} × 2^{−5} = 2^{3−5} = 2^{−2} , which is ^{1} ⁄_{4} .
(b) 3^{5} ÷ 3^{7}
Solution: 3^{5} ÷ 3^{7} = 3^{5−7} = 3^{−2} , which is ^{1} ⁄_{9} .
(c) (10^{2} )^{−3}
Solution: (10^{2} )^{−3} = 10^{(2 × (−3))} = 10^{−6} , which is ^{1} ⁄_{1,000,000} .
Click on questions to reveal their solutions
3. Fractional Powers
If a is positive , then the square root of a is a number which, when muliplied by itself,
gives a . Thus 3 is the square root of 9, since 3^{2} = 9.
We write 3 = √ 9 .
Note that, by definition, √ a × √ a = a .
This gives us a way of interpreting a ^{1⁄2} for, by Rule 1:
a ^{1⁄2} × a ^{1⁄2} = a ^{(1⁄2 + 1⁄2)} = a ^{1} = a = √ a × √ a
- so that a ^{1⁄2} = √ a .
The general rule is that, if a is a positive number and n is a positive integer, then:
- where n √a is the n th root of a . We can see this in general for, by Rule 3:
(a ^{1⁄n} )^{n} = a ^{1⁄n × n} = a ^{1} = a
Example 4:
(a) 100^{1⁄2} = √ 100 = 10
(b) 8^{1⁄3} = 3 √8 = 2
(c) 27^{5⁄3} = (27^{1⁄3} )^{5} = 3^{5} = 243
Note that in Example 4(c) we used Rule 3, i.e. (a ^{1⁄n} )^{m} = a ^{1⁄n × m} = (a ^{1⁄m} )^{n} , so:
(a ^{1⁄n} )^{m} = a ^{m⁄n} = (a ^{1⁄m} )^{n}
Quiz 1: To which of the following does (8^{5} )^{1⁄3} simplify?
(a) 8 Incorrect - please try again!
(b) 16 Incorrect - please try again!
(c) 24 Incorrect - please try again!
(d) 32 Correct - well done!
Explanation: Using Rule 3, we have:
(8^{5} )^{1⁄3} = (8^{1⁄3} )^{5} = (2)^{5} = 32
4. Use of the Rules of Simplification
In this section we shall demonstrate the use of the rules of powers to simplify more complicated expressions.
Example 5:
(a) [ (a ^{−3} )^{2⁄3} ] ^{1⁄2}
Beginning with the innermost bracket, we have, using Rule 3:
(a ^{−3} )^{2⁄3} = a ^{−3 × 2⁄3} = a ^{−2}
Then:
[ (a ^{−3} )^{2⁄3} ] ^{1⁄2} = [ a ^{−2} ] ^{1⁄2} = a ^{−2 × 1⁄2} = a ^{−1}
(b) [ (x ^{−1⁄4} )^{8} ] ^{2⁄3}
Beginning again with the innermost bracket, and using Rule 3, we have:
(x ^{−1⁄4} )^{8} = x ^{−1⁄4 × 8} = x ^{−2}
Now if we use Rule 3 again we have:
[ x ^{−2} ] ^{2⁄3} = x ^{−2 × 2⁄3} = x ^{−4⁄3}
(c) (x ^{1⁄2} )^{3} × (x ^{−1⁄3} )^{2}
We have:
(x ^{1⁄2} )^{3} × (x ^{−1⁄3} )^{2} = x ^{3⁄2} × x ^{−2⁄3}
- using Rule 3. Now we use Rule 1:
x ^{3⁄2} × x ^{−2⁄3} = x ^{3⁄2 − 2⁄3} = x ^{5⁄6}
(d) (4 √x ^{3} )^{2⁄3} × (5 √x ^{6} )^{5⁄12}
Starting with the first term:
4 √x ^{3} = (x ^{3} )^{1⁄4} = x ^{3⁄4}
Thus:
(4 √x ^{3} )^{2⁄3} = (x ^{3⁄4} )^{2⁄3} = x ^{3⁄4 × 2⁄3} = x ^{2⁄4} = x ^{1⁄2}
Similarly:
5 √x ^{6} = (x ^{6} )^{1⁄5} = x ^{6 × 1⁄5} = x ^{6⁄5}
so that:
(5 √x ^{6} )^{5⁄12} = (x ^{6⁄5} )^{5⁄12} = x ^{6⁄5 × 5⁄12} = x ^{1⁄2}
Finally we have:
(4 √x ^{3} )^{2⁄3} × (5 √x ^{6} )^{5⁄12} = x ^{1⁄2} × x ^{1⁄2} = x ^{1} = x
(e) (a ^{2} / b ^{3} )^{1⁄3} ×
(b ^{2} / a ^{3} )^{1⁄2}
The first term simplifies as follows.
( a ^{2} / b ^{3} ) ^{1⁄3} = (a ^{2} )^{1⁄3} / (b ^{3} )^{1⁄3} = a ^{2⁄3} / b = a ^{2⁄3} b ^{−1}
Treating the second term:
( b ^{2} / a ^{3} ) ^{1⁄2} = (b ^{2} )^{1⁄2} / (a ^{3} )^{1⁄2} = b / a ^{2⁄3} = b a ^{−3⁄2}
Thus:
( a ^{2} / b ^{3} ) ^{1⁄3} × ( b ^{2} / a ^{3} ) ^{1⁄2} = a ^{2⁄3} b ^{−1} × b a ^{−3⁄2} = a ^{2⁄3 − 3⁄2} = a ^{−5⁄6}
5. Quiz on Powers
Simplify the expressions and choose the solutions from the options given.
1. ( 3 √a ^{5} ) ^{1⁄2} × 6 √a ^{−5}
(a) 1
(b) a
(c) a ^{5⁄12}
(d) a ^{5⁄6}
2. ( a ^{3} / b ^{2} ) ^{1⁄2} ÷ ( b ^{3} / a ^{2} ) ^{−1⁄2}
(a) a ^{−1⁄2} b ^{−1⁄2}
(b) a ^{1⁄2} b ^{−1⁄2}
(c) a ^{−1⁄2} b ^{1⁄2}
(d) a ^{1⁄2} b ^{1⁄2}
3. (4 √b ^{3} )^{1⁄6} × 9 √b ^{−3} ÷ ( √ b ^{−7} )^{1⁄7}
(a) b ^{5⁄24}
(b) b ^{−5⁄24}
(c) b ^{7⁄24}
(d) b ^{−7⁄24}
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