1. Indices, surds, logarithms and exponential functions are covered. Key points include laws of indices, rationalizing surds, laws of logarithms including change of base, and the general forms of exponential functions as ax and ex. 2. Example questions demonstrate working with indices, simplifying surds, solving logarithmic and exponential equations using appropriate logarithm laws and change of variable techniques. 3. Solutions to example equations involve algebraic manipulation and setting logarithmic and exponential terms equal to solve for variables.

1. Indices, Surds &
Logarithms
Additional Math Notes

2. Indices
 54= 5 x 5 x 5 x 5 = 625
 5 is the base
 4 is the index or power

3. Laws of Indices
am
´ an
= am+n
am
¸ an
= am-n
(am
)n
= amn
am
´bm
= (a´b)m
am
¸ bm
= (
a
b
)m
a0
=1
a-m
=
1
am
(
a
b
)-m
= (
b
a
)m
a
1
m
= a
m
a
n
m
= ( a
m
)n
= an
m
(am
´ bn
)l
= aml
´bnl
Gotta Memorize these!!!

4. Example Questions
3x+y
´32x-y
= 3(x+y)+(2x-y)
= 33x
6x+y
¸ 6x-y
= 6(x+y)-(x-y)
= 62y
Q1 Q2

5. Example Questions
33
´ x3
= (3´ x)3
= (3x)3
x6
¸26
= (
x
2
)6
(215
)x
= 215x
3=7
=
1
37
(
16
7
)-3
= (
7
16
)3
Q3
Q4
Q5
Q6
Q7

6. Example Questions
1. Evaluate (18)
3
2
´(6)
-
1
2
´
1
27
= (2´32
)
3
2
´(2´3)
-
1
2
´
1
(33
)
1
2
= (2´32
)
3
2
´(2´3)
-
1
2
´(33
)
-
1
2
= 2
3
2
´33
´ 2
-
1
2
´3
-
1
2
´3
-
3
2
= 2
3
2
-
1
2
´3
3-
1
2
-
3
2
= 21
´31
= 6
2. Solve the equation 81n
= 9
81= 34
9 = 32
Hence, equation is (34
)n
= 32
34n
= 32
4n = 2
n =
1
2

7. Example Questions
3. Simplify
3n+1
´3n-1
9-n
=
3(n+1)+(n-1)
3-2n
=
3n+1+n-1
3-2n
= 32n-(-2n)
= 34n
4. Solve the equation 22x+1
-3(2x
)+1= 0
(22x
´ 21
)-3(2x
)+1= 0
Let 2x
be y
2y2
-3y +1= 0
(2y -1)(y -1) = 0
y =
1
2
, y =1
2x
=
1
2
® 2x
= 2-1
® x = -1
2x
=1® 2x
= 20
® x = 0

8. Example Questions
5. Solve the equation 9x
´ 22x
= 6
32x
´ 22x
= 6
(3´ 2)2x
= 6
62x
= 61
2x =1
x =
1
2
6. Solve the equation 81x
= 273x-5
34
= 33(3x-5)
34
= 39x-15
4 = 9x -15
9x =15+ 4
9x =19
x = 2
1
9

9. Surds
 surds is a subset of irrational numbers
 Eg √2, √3, √6 etc
 Multiplication of surds
 √a × √b = √ab
 Eg √5 × √7 = √30
 Addition and Subtraction of surds
 Eg 3√2 + 2√2 − 4√2 = 5√2 – 4√ 2 = √2
 Division of surds
 Eg √100 ÷ √25 = √(100÷25) = √4 = 2

10. Rationalizing Surds
Q2.
2 +1
11+3
=
( 2 +1)
( 11+3)
´
( 11-3)
( 11-3)
=
( 2 +1)( 11-3)
11-9
=
( 2 +1)( 11-3)
2
Q1.
2
3
=
2
3
´
3
3
=
2 3
3
Rationalize if there are
surds in the denominator
(a+b)(a-b) = a2 – b2

11. Example Questions
1. Simplify.
1
5
+ 20 + 125
=
1
5
*
5
5
+ 2´ 2´ 5 + 5´ 5´ 5
=
5
5
+ 2 5 + 5 5
= 5(
1
5
+ 2 + 5)
= 7
1
5
5
2. Simplify and express in the from a+b c
( 5 -2)2
= ( 5)2
- 2( 5)(2)+22
= 5- 4 5 + 4
= 9 - 4 5

12. Laws of Logarithm
If y=ax
, x is defined as the logarithm of y to the base a.
® x = loga y
1.loga xy = loga x + loga y
2.loga
x
y
= loga x - log a y
3.log(x)n
= nloga x
4.loga x
n
= loga x
1
n
=
1
n
loga x
Note:
- The log of any negative number to any base does not exist
eg. log5(-10) does not exist
- The log of 1 to any base is zero
eg. log31= 0
- logx x =1

13. Logarithm
 Logarithms to base 10 are called common logarithms
 Change of base
log10 a ®loga or lga
loga x =
logb x
logb a
loga x =
1
logx a
loga b =
logb
loga
How to calculate?

14. Example Questions
1. Solve the equation 2x
´3x
= 5x+1
2x
´3x
= 5x+1
6x
= 5x+1
xlog10 6 = (x +1)log10 5
x +1
x
=
log10 6
log10 5
=
0.7782
0.6990
=1.113
x +1=1.113x
(1.113-1)x =1
x = 8.85 (2d.p)
2. Simplify log3 2+ log3 5+ log3 20 - log3 25
= log3(
2´5´20
25
)
= log3(
200
25
)
= log3 8
= log3 23
= 3log3 2

15. Example Questions
3. Solve the equation
(log5 x)2
-3log5 x + 2 = 0
Let log5 x = y
y2
-3y + 2 = 0
(y - 2)(y -1) = 0
y = 2, y =1
log5 x = 2
x = 52
= 25
log5 x =1
x = 51
= 5
4. Solve the equation log4 x - logx 8 =
1
2
log4 x -
log4 8
log4 x
=
1
2
log4 x -
1.5
log4 x
=
1
2
let log4 x be y
y-
1.5
y
=
1
2
´ y : y2
-1.5 =
1
2
y
y2
-
1
2
y -1.5 = 0
´ 2: 2y2
- y -3= 0
(2y -3)(y+1) = 0
2y = 3, y = -1
y =
3
2
log4 x =
3
2
x = 4
3
2
x = 8
y = -1
log4 x = -1
x = 4-1
x =
1
4

16. Example Questions
5. Solve the equation
log9[log2 (4x -16)]= log16 4
log9[log2 (4x -16)]=
1
2
log9[log2 (4x -16)]= log3 9
log2 (4x -16) = 3
log2 (4x -16) = log2 8
4x -16 = 8
4x = 24
x = 6
6. Solve the simultaneous equations
log2x y = 2 ® eqn1
logx 4y = 6 ® eqn2
eqn1: y = (2x)2
y = 4x2
® eqn3
sub eqn3 into eqn2
logx 4(4x2
) = 6
logx 16x2
= 6
logx 16x2
= logx x6
16x2
= x6
16x2
- x6
= 0
x2
(16 - x4
) = 0
x2
= 0(rejected),16 - x4
= 0
16 = x4
x = 2,-2(rejected)

17. Exponential Function
 General form is ax where a is a positive constant and x is
a variable
 Important exponential functions
 10x
 ex

18. Example Questions
1. Solve the equation
e2ln x
+ lne2x
= 8
e2ln x
+ 2x = 8
e2ln x
= 8- 2x
2ln x = ln(8- 2x)
ln x2
= ln(8- 2x)
x2
= 8- 2x
x2
-8+ 2x = 0
(x + 4)(x - 2) = 0
x = -4(rejected), x = 2
2. Solve the equation
10x
= e2x+1
ln10x
= lne2x+!
xln10 = 2x +1
xln10 - 2x =1
x(ln10 - 2) =1
x =
1
ln10 - 2
x = 3.30

19. Example Questions
3. Solve the equation
2e2x+!
= ex+1
+15e
2e2x
*e = ex
*e+15e
let ex
be y
2ey2
= ey +15e
2ey2
-ey -15e = 0
e(2y2
- y -15) = 0
2y2
- y -15 = 0
(2y + 5)(y -3) = 0
y = -
5
2
, y = 3
ex
= -
5
2
(rejected)
ex
= 3
x =1.10
4. Solve the equation
ex
= 2e
x
2
+15
ex
= 2(ex
)
1
2
+15
let ex
be y
y=2y
1
2
+15
y - 2y
1
2
-15 = 0
(y
1
2
- 5)(y
1
2
+3) = 0
y
1
2
= 5,y
1
2
= -3(rejected)
y = 25
ex
= 25
x = ln25 = 3.22
5. Solve the equation
e3x-1
=148
lne3x-1
= ln148
3x -1= 5
3x = 6
x = 2