5891862.ppt

MULTIPLYING INDICES
am x an
= am + n

MULTIPLYING INDICES
a2 x a3
= a2 + 3
a5

DIVIDING INDICES
(m greater than n)
m > n
am ÷ an
= am - n

DIVIDING INDICES
a4 ÷ a2
= a4 - 2
a2

DIVIDING INDICES
(m less than n)
m < n
am ÷ an
= 1/ an - m

DIVIDING INDICES
a2 ÷ a4
= 1/a4 - 2
1/a2

DIVIDING INDICES (ALTERNATIVE)
a2 ÷ a4
= a -2
1/a2 Same answer

INDICES IN BRACKETS
(am)n
= amn

INDICES IN BRACKETS
(a2)3
= a6

REMEMBER
Any number to the power 0 =1
90 = 1
1000 = 1

WORKED EXAMPLE
3a2b3 x 2a4b
Separate the terms
3 x 2 = 6
a2 x a4 = a6
b3 x b = b4
Answer = 6a6b4

WORKED EXAMPLE
(2c3d2)2
All the terms inside the brackets are squared
22 x c 3x2 x d2x2
= 22c6d4

WORKED EXAMPLE
a) Show that 43/2 = 8
43/2 means the square root of 4
cubed (√43)
The square root of 4 = 2,
23 = 8

WORKED EXAMPLE
b), solve the equation 4x = 84
43/2 = 8 so 84 = 4 4x3/2
x = 4x3/2 = 6

WORKED EXAMPLE
Evaluate (1/3)-3
(1/3)-3 is the same as (3/1)3
33 = 27

INDICES AND LOGARITHMS
N = ax logaN = x
4 = 22 log24 = 2
8 = 23 log28 = 3

100 = 102 log10100 = 2
1000 = 103 log101000 = 3

log ab
= log a + log b

log10 8*5
log 108 + log105
0.903 + 0.70
= 1.60

log a/b
= log a - log b

log10 8/5
log 108 - log105
0.903 - 0.70
= 0.203

log xn
n.log x

NATURAL LOGARITHMS
The natural logarithm is the logarithm to the base
e
e is Euler's number, the base of natural
logarithms,
e approximates to 2.718
also known as Napier's constant

SIMULTANEOUS EQUATIONS ( BY ELIMINATION)
1, 2x - y = 2
2, x + y = 7
Add 1, and 2, (because there is a +y and a – y)
3x = 9
x = 3 substitute for x in 1,
6 – y = 2
y = 4

1, 2x + y = 7
2, x + y = 4
Subtract 2, from 1, (because there are two + y’s)
x = 3
Substitute for x in 1,
y = 1

1, 3x + y = 9
2, 2x +2y = 10
Multiply 1, by 2
3, 6x + 2y = 18
Subtract 2, from 3,
4x = 8
X = 2
Y = 3

SIMULTANEOUS EQUATIONS ( BY SUBSTITUTION
1, y = 5x -3
2, y = 3x + 7
5x – 3 = 3x + 7
(rearrange)
5x – 3x = 7 + 3
2x = 10 x = 5
(substitute in 1)
y = (5x5) – 3 = 25 - 3 = 22

SIMULTANEOUS EQUATIONS ( BY SUBSTITUTION)
1,2x + y = 7
2, x + y = 4
x = 4 – y
Substitute in 1,
2(4 - y) + y = 7
8 -2y + y = 7
8 – y = 7
Y = 1 (substitute in 2,)
1 + x = 4
X = 3

SIMULTANEOUS EQUATIONS ( BY GRAPHICAL
INTERCEPTION)

WORDED SIMULTANEOUS EQUATION
Bill has more money than Mary. If Bill gave
Mary £20, they would have the same
amount. While if Mary gave Bill £22, Bill
would then have twice as much as
Mary. How much does each one actually
have?

If Bill gave Mary £20, they would
have the same amount."
Algebraically:
1) B − 20 = M + 20.
Or B = M + 40

While if Mary gave Bill £22, Bill would
then have twice as much
as Mary."
Algebraically:
B + 22 = 2(M − 22).
B = 2M – 44 – 22
B = 2M - 66

Now we have two equations
B = M + 40 and
B = 2M – 66 so
2M – 66 = M + 40
2M -M = 40 + 66
M = 106
Mary has £106
B = M + 40
Bill has £146

The effort (E) required to raise a load (W) using a
certain hoisting mechanism is related by the
equation: E = aW+b. During tests it was found that
an effort of 4.5 N would raise a load of 15kg and an
effort of 10N would raise a load of 30kg. Calculate
the constants a and b for the machine equation and
hence determine the effort required to raise a
S.W.L. (Safe Working Load) of 100 kg.

E = aW+b. 1, 4.5 = 15a + b and
2, 10 = 30a + b
5.5 = 15a
a = 5.5/15 = 0.366
Substitute in 1,
4.5 = (15 x 0.366) + b
4.5 = 5.5 + b
B = -1
E = (100 x 0.366) + b
E = 36.6 -1 = 35.6N

5891862.ppt

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5891862.ppt