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.
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!!!
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?
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
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