Indices, Surds & Logarithms
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
a1
m = am
an
m = ( am )n = a nm
(am ´ bn )l = aml ´bnlGotta Memorize these!!!
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
= 23
2 ´33 ´ 2-
1
2 ´3-
1
2 ´ 3-
3
2
= 23
2-
1
2 ´33-
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 22 x+1 - 3(2x )+1= 0
(22 x ´ 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 ´ 22 x = 6
32 x ´ 22 x = 6
(3´ 2)2 x = 6
62 x = 61
2x =1
x =1
2
6. Solve the equation 81x = 273x-5
34 = 33(3x-5)
34 = 39 x-15
4 = 9x -15
9x =15+ 4
9x =19
x = 21
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)
= 71
55
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=a x, x is defined as the logarithm of y to the base a.
® x = loga y
1. loga xy = loga x + loga y
2. logax
y= loga x - log a y
3. log(x)n = n loga x
4. loga xn = loga x1
n =1
nloga 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
- logxx =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
1. Solve the equation 2 x ´3x = 5x+1
2x ´3x = 5x+1
6x = 5x+1
x log10 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 log4x be y
y-1.5
y=
1
2
´ y : y2 -1.5 =1
2y
y2 -1
2y-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 = 43
2
x = 8
y = -1
log4 x = -1
x = 4-1
x =1
4
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
log2 xy = 2® eqn1
logx 4y = 6® eqn2
eqn1: y = (2x)2
y = 4x2 ® eqn3
sub eqn3 into eqn2
logx 4(4x2 ) = 6
logx16x2 = 6
logx16x2 = 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
1. Solve the equation
e2 ln x + lne2 x = 8
e2 ln x + 2x = 8
e2 ln x = 8- 2x
2 ln 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 = e2 x+1
ln10x = lne2 x+!
x ln10 = 2x +1
x ln10 - 2x =1
x(ln10 - 2) =1
x =1
ln10 - 2
x = 3.30
Example Questions 3. Solve the equation
2e2 x+! = ex+1 +15e
2e2 x *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 = 2ex
2 +15
ex = 2(ex )1
2 +15
let ex be y
y=2y1
2 +15
y- 2y1
2 -15 = 0
(y1
2 - 5)(y1
2 + 3) = 0
y1
2 = 5, y1
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