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Laws of indices

Laws of indices. www.mathschampion.co.uk. Multiplying indices. a m x a n = a m + n. Multiplying indices. a 2 x a 3 = a 2 + 3 a 5. dividing indices ( m greater than n ). m > n a m ÷ a n = a m - n. dividing indices. a 4 ÷ a 2 = a 4 - 2 a 2.

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Laws of indices

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  1. Laws of indices www.mathschampion.co.uk

  2. Multiplying indices am x an = am+ n

  3. Multiplying indices a2 x a3 = a2+ 3 a5

  4. dividing indices (m greater than n) m > n am ÷ an = am- n

  5. dividing indices a4 ÷ a2 = a4- 2 a2

  6. dividing indices (m less than n) m < n am ÷ an = 1/ an- m

  7. dividing indices a2 ÷ a4 = 1/a4- 2 1/a2

  8. dividing indices (ALTERNATIVE) a2 ÷ a4 = a -2 1/a2 Same answer

  9. Indices in brackets (am)n = amn

  10. Indices in brackets (a2)3 = a6

  11. Remember • Any number to the power 0 =1 • 90 = 1 • 1000 = 1

  12. Worked example 3a2b3x 2a4b Separate the terms 3 x 2 = 6 a2 x a4 = a6 b3 x b = b4 Answer = 6a6b4

  13. Worked example (2c3d2)2 All the terms inside the brackets are squared 22 x c 3x2 x d2x2 = 22c6d4

  14. 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

  15. Worked example • b), solve the equation 4x= 84 • 43/2 = 8 so 84 = 4 4x3/2 • x = 4x3/2 = 6

  16. Worked example Evaluate (1/3)-3 (1/3)-3 is the same as (3/1)3 33 = 27

  17. Indices and logarithms N = axlogaN = x 4 = 22 log24 = 2 8 = 23 log28 = 3

  18. Indices and logarithms 100 = 102 log10100 = 2 1000 = 103 log101000 = 3

  19. Indices and logarithms • log ab • = log a + log b

  20. Indices and logarithms • log10 8*5 • log 108 + log105 • 0.903 + 0.70 • = 1.60

  21. Indices and logarithms • log a/b • = log a - log b

  22. Indices and logarithms • log108/5 • log 108 - log105 • 0.903 - 0.70 • = 0.203

  23. Indices and logarithms • log xn • n.log x

  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

  25. 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

  26. 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

  27. 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

  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

  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

  30. Simultaneous equations ( by graphical interception)

  31. 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?

  32. Worded simultaneous equation • If Bill gave Mary £20, they would have the same amount." • Algebraically: • B− 20 = M+ 20. • Or B = M + 40

  33. 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

  34. 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

  35. 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.

  36. 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

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