Transcript Laws of indices - Maths Champion

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MULTIPLYING INDICES
m
a
=
x
n
a
m
+
n
a
MULTIPLYING INDICES
2
a
3
a
x
2
+
3
= a
5
a
DIVIDING INDICES
(m greater than n)
m>n
m
n
a ÷ a
=
m
n
a
DIVIDING INDICES
÷
4
2
= a
2
a
4
a
2
a
DIVIDING INDICES
(m less than n)
m<n
m
n
a ÷ a
= 1/
n
m
a
DIVIDING INDICES
÷
4
2
= 1/a
2
1/a
2
a
4
a
DIVIDING INDICES (ALTERNATIVE)
÷
-2
= a
2
1/a
2
a
4
a
Same answer
INDICES IN BRACKETS
m
n
(a )
=
mn
a
INDICES IN BRACKETS
2
3
(a )
=
6
a
REMEMBER
Any number to the power 0 =1
90 = 1
1000 = 1
WORKED EXAMPLE
3a2b3 x 2a4b
Separate the terms
3x2=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
INDICES AND LOGARITHMS
100 = 102
log10100 = 2
1000 = 103
log101000 = 3
INDICES AND LOGARITHMS
log ab
= log a + log b
INDICES AND LOGARITHMS
log10 8*5
log 108 + log105
0.903 + 0.70
= 1.60
INDICES AND LOGARITHMS
log a/b
= log a - log b
INDICES AND LOGARITHMS
log10 8/5
log 108 - log105
0.903 - 0.70
= 0.203
INDICES AND LOGARITHMS
log
n
x
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
SIMULTANEOUS EQUATIONS ( BY ELIMINATION)
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
SIMULTANEOUS EQUATIONS ( BY ELIMINATION)
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?
WORDED SIMULTANEOUS EQUATION
If Bill gave Mary £20, they would
have the same amount."
Algebraically:
1) B − 20 = M + 20.
Or B = M + 40
WORDED SIMULTANEOUS EQUATION
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
WORDED SIMULTANEOUS EQUATION
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
WORDED SIMULTANEOUS EQUATION
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.
WORDED SIMULTANEOUS EQUATION
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