1. LAWS OF INDICES www.mathschampion.co.uk

2. MULTIPLYING INDICES a m x a n = a m + n

3. MULTIPLYING INDICES a 2 x a 3 = a 2 + 3 a5a5

4. DIVIDING INDICES (m greater than n) m > n a m ÷ a n = a m - n

5. DIVIDING INDICES a 4 ÷ a 2 = a 4 - 2 a2a2

6. DIVIDING INDICES (m less than n) m < n a m ÷ a n = 1/ a n - m

7. DIVIDING INDICES a 2 ÷ a 4 = 1/a 4 - 2 1/a 2

8. DIVIDING INDICES (ALTERNATIVE) a 2 ÷ a 4 = a -2 1/a 2 Same answer

9. INDICES IN BRACKETS (a m ) n = a mn

10. INDICES IN BRACKETS (a 2 ) 3 = a 6

11. REMEMBER Any number to the power 0 =1 9 0 = 1 100 0 = 1

12. 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

13. 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

14. 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

15. WORKED EXAMPLE b), solve the equation 4 x = 8 4 4 3/2 = 8 so 8 4 = 4 4x3/2 x = 4x3/2 = 6

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

17. INDICES AND LOGARITHMS N = a x log a N = x 4 = 2 2 log 2 4 = 2 8 = 2 3 log 2 8 = 3

18. INDICES AND LOGARITHMS 100 = 10 2 log 10 100 = 2 1000 = 10 3 log 10 1000 = 3

19. INDICES AND LOGARITHMS log ab = log a + log b

20. INDICES AND LOGARITHMS log 10 8*5 log 10 8 + log 10 5 0.903 + 0.70 = 1.60

21. INDICES AND LOGARITHMS log a/b = log a - log b

22. INDICES AND LOGARITHMS log 10 8/5 log 10 8 - log 10 5 0.903 - 0.70 = 0.203

23. INDICES AND LOGARITHMS log x n 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: 1) 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