Basic algebra 65
(2)
a m
a n = a m−n For example,
c 5
c 2 = c 5−2 = c 3
(3) (a m ) n = a mn For example,
d 2 3 = d 2×3 = d 6
(4) a
m
n =
n
√
a m For example, x
4
3 =
3
√
x 4
(5) a −n =
1
a n
For example, 3 −2 =
1
3 2 =
1
9
(6) a 0 = 1
For example, 17 0 = 1
Here are some worked examples to demonstrate these
laws of indices.
Problem 17. Simplify a 2 b 3 c × ab 2 c 5
a
2 b
3 c × ab
2 c
5
= a
2
× b
3
× c × a × b
2
× c
5
= a
2
× b
3
× c
1
× a
1
× b
2
× c
5
Grouping together like terms gives
a 2 × a 1 × b 3 × b 2 × c 1 × c 5
Using law (1) of indices gives
a 2+1 × b 3+2 × c 1+5 = a 3 × b 5 × c 6
i.e.
a 2 b 3 c × ab 2 c 5 = a 3 b 5 c 6
Problem 18. Simplify a
1
3 b
3
2 c −2 × a
1
6 b
1
2 c
Using law (1) of indices,
a
1
3 b
3
2 c
−2
× a
1
6 b
1
2 c = a
1
3 +
1
6 × b
3
2 +
1
2 × c
−2+1
= a
1
2 b
2 c
−1
Problem 19. Simplify
x 5 y 2 z
x 2 y z 3
x 5 y 2 z
x 2 yz 3 =
x 5 × y 2 × z
x 2 × y × z 3
=
x 5
x 2 ×
y 2
y 1 ×
z
z 3
= x
5−2
× y
2−1
× z
1−3 by law (2) of indices
= x
3
× y
1
× z
−2
= x
3 y z
−2 or
x 3 y
z 2
Problem 20. Simplify
a
3 b
2 c
4
abc −2 and evaluate when
a = 3, b =
1
4
and c = 2
Using law (2) of indices,
a 3
a
= a
3−1
= a
2
,
b 2
b
= b
2−1
= b and
c 4
c −2 = c
4− −2
= c
6
Thus,
a
3 b
2 c
4
abc −2 = a 2 bc
6
When a = 3, b =
1
4
and c = 2,
a 2 bc 6 = (3) 2
1
4
(2) 6 = (9)
1
4
(64) = 144
Problem 21. Simplify ( p 3 ) 2 (q 2 ) 4
Using law (3) of indices gives
( p
3
)
2
(q
2
)
4
= p
3×2
× q
2×4
= p
6 q
8
Problem 22. Simplify
(mn 2 ) 3
(m 1/2 n 1/4 ) 4
The brackets indicate that each letter in the bracket must
be raised to the power outside. Using law (3) of indices
gives
(mn 2 ) 3
(m 1/2 n 1/4 ) 4 =
m 1×3 n 2×3
m (1/2)×4 n (1/4)×4 =
m 3 n 6
m 2 n 1
Using law (2) of indices gives
m 3 n 6
m 2 n 1 = m
3−2 n
6−1
= mn
5
Problem 23. Simplify (a 3 b)(a −4 b −2 ), expressing
the answer with positive indices only
Using law (1) of indices gives a 3+−4 b 1+−2 = a −1 b −1
Using law (5) of indices gives a −1 b −1 =
1
a +1 b +1 =
1
ab
Problem 24. Simplify
d 2 e 2 f 1/2
(d 3/2 ef 5/2 ) 2 expressing
the answer with positive indices only
(2)
a m
a n = a m−n For example,
c 5
c 2 = c 5−2 = c 3
(3) (a m ) n = a mn For example,
d 2 3 = d 2×3 = d 6
(4) a
m
n =
n
√
a m For example, x
4
3 =
3
√
x 4
(5) a −n =
1
a n
For example, 3 −2 =
1
3 2 =
1
9
(6) a 0 = 1
For example, 17 0 = 1
Here are some worked examples to demonstrate these
laws of indices.
Problem 17. Simplify a 2 b 3 c × ab 2 c 5
a
2 b
3 c × ab
2 c
5
= a
2
× b
3
× c × a × b
2
× c
5
= a
2
× b
3
× c
1
× a
1
× b
2
× c
5
Grouping together like terms gives
a 2 × a 1 × b 3 × b 2 × c 1 × c 5
Using law (1) of indices gives
a 2+1 × b 3+2 × c 1+5 = a 3 × b 5 × c 6
i.e.
a 2 b 3 c × ab 2 c 5 = a 3 b 5 c 6
Problem 18. Simplify a
1
3 b
3
2 c −2 × a
1
6 b
1
2 c
Using law (1) of indices,
a
1
3 b
3
2 c
−2
× a
1
6 b
1
2 c = a
1
3 +
1
6 × b
3
2 +
1
2 × c
−2+1
= a
1
2 b
2 c
−1
Problem 19. Simplify
x 5 y 2 z
x 2 y z 3
x 5 y 2 z
x 2 yz 3 =
x 5 × y 2 × z
x 2 × y × z 3
=
x 5
x 2 ×
y 2
y 1 ×
z
z 3
= x
5−2
× y
2−1
× z
1−3 by law (2) of indices
= x
3
× y
1
× z
−2
= x
3 y z
−2 or
x 3 y
z 2
Problem 20. Simplify
a
3 b
2 c
4
abc −2 and evaluate when
a = 3, b =
1
4
and c = 2
Using law (2) of indices,
a 3
a
= a
3−1
= a
2
,
b 2
b
= b
2−1
= b and
c 4
c −2 = c
4− −2
= c
6
Thus,
a
3 b
2 c
4
abc −2 = a 2 bc
6
When a = 3, b =
1
4
and c = 2,
a 2 bc 6 = (3) 2
1
4
(2) 6 = (9)
1
4
(64) = 144
Problem 21. Simplify ( p 3 ) 2 (q 2 ) 4
Using law (3) of indices gives
( p
3
)
2
(q
2
)
4
= p
3×2
× q
2×4
= p
6 q
8
Problem 22. Simplify
(mn 2 ) 3
(m 1/2 n 1/4 ) 4
The brackets indicate that each letter in the bracket must
be raised to the power outside. Using law (3) of indices
gives
(mn 2 ) 3
(m 1/2 n 1/4 ) 4 =
m 1×3 n 2×3
m (1/2)×4 n (1/4)×4 =
m 3 n 6
m 2 n 1
Using law (2) of indices gives
m 3 n 6
m 2 n 1 = m
3−2 n
6−1
= mn
5
Problem 23. Simplify (a 3 b)(a −4 b −2 ), expressing
the answer with positive indices only
Using law (1) of indices gives a 3+−4 b 1+−2 = a −1 b −1
Using law (5) of indices gives a −1 b −1 =
1
a +1 b +1 =
1
ab
Problem 24. Simplify
d 2 e 2 f 1/2
(d 3/2 ef 5/2 ) 2 expressing
the answer with positive indices only