indice-ppt.ppt

Indices, Surds &
Logarithms
Additional Math Notes

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

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!!!

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

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

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

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

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

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

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

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

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?

Example Questions
´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

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

Example Questions
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)

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

Example Questions
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
10x
= e2x+1
ln10x
= lne2x+!
xln10 = 2x +1
xln10 - 2x =1
x(ln10 - 2) =1
x =
1
ln10 - 2
x = 3.30

Example Questions
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
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
e3x-1
=148
lne3x-1
= ln148
3x -1= 5
3x = 6
x = 2

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