LAWS OF INDICES www.mathschampion.co.uk
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- Slide 1
- LAWS OF INDICES www.mathschampion.co.uk
- Slide 2
- MULTIPLYING INDICES a m x a n = a m + n
- Slide 3
- MULTIPLYING INDICES a 2 x a 3 = a 2 + 3 a5a5
- Slide 4
- DIVIDING INDICES (m greater than n) m > n a m a n = a m - n
- Slide 5
- DIVIDING INDICES a 4 a 2 = a 4 - 2 a2a2
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- DIVIDING INDICES (m less than n) m < n a m a n = 1/ a n - m
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- DIVIDING INDICES a 2 a 4 = 1/a 4 - 2 1/a 2
- Slide 8
- DIVIDING INDICES (ALTERNATIVE) a 2 a 4 = a -2 1/a 2 Same answer
- Slide 9
- INDICES IN BRACKETS (a m ) n = a mn
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- INDICES IN BRACKETS (a 2 ) 3 = a 6
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- REMEMBER Any number to the power 0 =1 9 0 = 1 100 0 = 1
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- WORKED EXAMPLE 3a 2 b 3 x 2a 4 b Separate the terms 3 x 2 = 6 a 2 x a 4 = a 6 b 3 x b = b 4 Answer = 6a 6 b 4
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- WORKED EXAMPLE (2c 3 d 2 ) 2 All the terms inside the brackets are squared 2 2 x c 3x2 x d 2x2 = 2 2 c 6 d 4
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- WORKED EXAMPLE a) Show that 4 3/2 = 8 4 3/2 means the square root of 4 cubed (4 3 ) The square root of 4 = 2, 2 3 = 8
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- WORKED EXAMPLE b), solve the equation 4 x = 8 4 4 3/2 = 8 so 8 4 = 4 4x3/2 x = 4x3/2 = 6
- Slide 16
- WORKED EXAMPLE Evaluate (1/3) -3 (1/3) -3 is the same as (3/1) 3 3 3 = 27
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- INDICES AND LOGARITHMS N = a x log a N = x 4 = 2 2 log 2 4 = 2 8 = 2 3 log 2 8 = 3
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- INDICES AND LOGARITHMS 100 = 10 2 log 10 100 = 2 1000 = 10 3 log 10 1000 = 3
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- INDICES AND LOGARITHMS log ab = log a + log b
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- INDICES AND LOGARITHMS log 10 8*5 log 10 8 + log 10 5 0.903 + 0.70 = 1.60
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- INDICES AND LOGARITHMS log a/b = log a - log b
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- INDICES AND LOGARITHMS log 10 8/5 log 10 8 - log 10 5 0.903 - 0.70 = 0.203
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- INDICES AND LOGARITHMS log x n n.log x
- Slide 24
- 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
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- 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
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- SIMULTANEOUS EQUATIONS ( BY ELIMINATION) 1, 2x + y = 7 2, x + y = 4 Subtract 2, from 1, (because there are two + ys) x = 3 Substitute for x in 1, y = 1
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- 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
- Slide 28
- 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
- Slide 29
- 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
- Slide 30
- SIMULTANEOUS EQUATIONS ( BY GRAPHICAL INTERCEPTION)
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- 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?
- Slide 32
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