Topic 3: Equations with exponents
Before starting to solve equations containing exponents, it is important to think about
opposite operations just like in the Equations Topic.
You will recall the opposites below:
Operation
Addition +
Opposite
Operation
Subtraction -
Subtraction -
Addition +
Multiplication x
Division ÷
Division ÷
Multiplication x
Squaring n 2
Square root n
Square root
n
Squaring n 2
Revision example:
2 x 2 + 3 = 11
2 x + 3 − 3 = 11 − 3 → take 3 from both sides
2 x2 = 8
2 x2 8
= → divide both sides by 2
2
2
x2 = 4
2
x 2 = ± 4 → take square root of both sides
x = ±2
There are two solutions here
because both 2 and -2 satisfy
the equation. This always
occurs with even roots.
2 x 2 + 3 = 11
2 x 2 = 11 − 3 → +3 becomes -3 on the RHS
2 x2 = 8
8
x 2 = → ×2 becomes ÷ 2 on the RHS
2
2
x =4
x = ± 4 →2 becomes
on the RHS
x = ±2
Centre for Teaching and Learning | Academic Practice | Academic Skills
T +61 2 6626 9262 E [email protected] W www.scu.edu.au/teachinglearning
Page 1
[last edited on 13 July 2015] CRICOS Provider: 01241G
Centre for Teaching and Learning
Numeracy
This list can be expanded to cubes, cube roots, to the 4, forth root, etc.
Operation
Cubing ( )
3
th
To the 4 power
th
To the 5 power
Opposite Operation
Cube root ( 3
) or
Forth root ( 4
) or
Fifth root (
5
6
Sixth root (
th
Seventh root ( 7
etc.
To the 6 power
To the 7 power
etc.
2
3
3
To the power
5
To the power
1
4
1
5
) or
th
1
3
1
6
) or
) or
1
7
3
2
5
To the power
3
To the power
Examples: Solve for x
Both sides method
x 4 = 256
4
4
4
x = 256 → take the fourth root of both sides
x = ±4
Short -cut Method
4
x = 256
x = 4 256 →
x = ±4
4
becomes
4
becomes
6
on the RHS
There are two solutions here
because both 4 and -4 satisfy
the equation. This always
occurs with even roots
6
x =3
( x)
6
6
= 36 raise both sides to the power of 6
x = 729
35 x −1 = 5
3 5 x − 1 + 1 = 5 + 1 → add 1 to both sides
35 x = 6
35 x 6
= → divide both sides by 3
3
3
5
x =2
( x)
5
5
= 25 → raise both side to the power of 5
x = 32
6
x =3
x = 36 →
x = 729
6
on the RHS
35 x −1 = 5
3 5 x = 5 + 1 → - 1 becomes + 1 on the RHS
35 x = 6
6
5
x = → × by 3 becomes ÷ by 3 on the RHS
3
5
x =2
5
x = 25 → 5 becomes ( ) on the RHS
x = 32
Centre for Teaching and Learning | Academic Practice | Academic Skills
T +61 2 6626 9262 E [email protected] W www.scu.edu.au/teachinglearning
Page 2
[last edited on 13 July 2015] CRICOS Provider: 01241G
Centre for Teaching and Learning
Numeracy
4 x −2 = 40
4 x −2 = 40
10
4 1 x −2 40
= 1 → divide both sides by 4
41
4
1
−2
x = 10 → raise both sides to the −
2
(x
−
= 10
1
x= 1
10 2
1
x=
10
x = ±0.3162 ( to 4 d.p.)
2
( 2 x − 1)
2
( 2 x − 1)
= 49
( 2 x − 1)
2
−
1
2
on the RHS
= 49
2 x − 1 = 49 → 2 becomes
on the RHS
2 x − 1 = ±7
2 x = 7 + 1 or − 7 + 1 → − 1 becomes + 1 on the RHS
2 x = 8 or − 6
−6
8
x = or
→ × 2 becomes ÷ 2 on the RHS
2
2
x = 4 or − 3
= 49 → take square root of both sides
2 x − 1 = ±7
2 x − 1 + 1 = 7 + 1 or − 7 + 1 → add 1 to both sides
2 x = 8 or − 6
−6
2x 8
= or
→ divide both sides by 2
2 2
2
x = 4 or − 3
In the question below, the opposite of raising a
number to a power is raising it to the reciprocal
power. Remember a number multiplied by its
reciprocal gives 1.
1
x = 10 2 → −2 becomes
1
x=
10
x = ±0.3162 ( to 4 d.p.)
1
−
2
1
−2 − 2
)
10
40
→ × 4 becomes ÷ 4 on the RHS
41
x −2 = 10
x −2 =
In the question below, the fraction exponent
becomes the reciprocal on the other side of the
equation.
2
2
x 3 = 50
x 3 = 50
3
3
3
 23  2
2
x
=
50
 
 
x = 503
x = 353.55
x = 50 2
x = 503
x = 353.55
It is also important to be able to rearrange formulae that contain powers.
Rearrange the formulae below to make the pronumeral in brackets the subject.
A = π r 2 [r ]
A π 1r 2
=
→ divide both sides by π
1
π
A
A
π
π
A
π
π
= r 2 → take the square root of both sides
= r2
r=
A = π r 2 [r ]
A 2
= r → ×π becomes ÷ π on LHS
π
= r →2 becomes
r=
on the LHS
A
π
A
π
Centre for Teaching and Learning | Academic Practice | Academic Skills
T +61 2 6626 9262 E [email protected] W www.scu.edu.au/teachinglearning
Page 3
[last edited on 13 July 2015] CRICOS Provider: 01241G
Centre for Teaching and Learning
Numeracy
4
4
V = π r 3 [ r ] The first step is to remove the fraction
V = π r 3 [ r ] The first step is to remove the fraction
3
3
4 3
1
3
V
4
=
π r 3 → ÷3 becomes × 3 on LHS
3V = 1 π r × 3 → Multiple both sides by 3
3
3V
= r 3 → × 4π becomes ÷ 4π on LHS
3V = 4π r 3
4
π
3V 4π r 3
3V
=
→ Divide both sides by 4π
3
= r → 3 becomes 3 on the LHS
4π
4π
4π
3V
= r3
3V
4π
r=3
4π
3V 3 3
3
= r → take cube root of both sides
4π
3V
r=3
4π
1
f =
[C ]
2π LC
1
f × LC =
× LC → multiply both sides by LC to get C into the numerator
2π LC
f × LC
1
=
→ divide both sides by f
f
2π f
(
2
 1 
=
 → square both sides
 2π f 
1
LC = 2 2
4π f
LC
1
= 2 2 → divide both sides by L
L
4π f L
1
C= 2 2
4π f L
LC
)
2
Or
1
[C ]
2π LC
1
f × LC =
→ ÷ LC becomes × LC on the LHS
2π
1
LC =
→ × f becomes ÷ f on the RHS
2π f
f =
2
 1 
LC = 
becomes 2 on the RHS
 →
2
f
π


1
LC = 2 2
4π f
1
C = 2 2 → × L becomes ÷ L on the RHS
4π f L
Video ‘Equations containing exponents’
Centre for Teaching and Learning | Academic Practice | Academic Skills
T +61 2 6626 9262 E [email protected] W www.scu.edu.au/teachinglearning
Page 4
[last edited on 13 July 2015] CRICOS Provider: 01241G
Centre for Teaching and Learning
Numeracy
Activity
1.
Solve for x :
(a)
x3 = 27
(b)
x5 + 1 = 244
(c)
34 x = 6
(d)
3
(e)
12
(f)
4 x + 31 = 63
4
x +1 = 5
2
x =2
(g)
5x + 4 = 9
(h)
7( x + 1) = 119
(i)
x −4 = 20 (use calc.)
(j)
x −1
= 11
4
(k)
4x−5 = 5
(l)
3x 2 + 7 = 19
(m)
3x4 + 256 = 1024
2.
Rearrange these formulae for the pronumeral in brackets
(a)
A = 4π r 2
(d)
F=
(g)
Z = R 2 + ω 2 L2
1 2
mv
2
1
[r ]
(b)
V = π r 2h
[r ]
(c)
c2 = a 2 + b2
[b]
[v]
(e)
Q = SLd 2
[d ]
(f)
v 2 = u 2 + 2as
[u ]
(h)
R=
2GM
c2
[c]
(i)
P=
[ L]
π 2 EI
L2
[ L]
The next questions are challenging, take care!
(j)
r3
T = 2π
GM
(m)
A = P(1 + i) n
[r ]
(k)

v2 
F = m g − 
R

[v]
(l)
R=
π r 4 ( p2 − p1 )
8nL
[r ]
[i ]
Centre for Teaching and Learning | Academic Practice | Academic Skills
T +61 2 6626 9262 E [email protected] W www.scu.edu.au/teachinglearning
Page 5
[last edited on 13 July 2015] CRICOS Provider: 01241G