2. HIGHER MATHEMATICS Unit 3 - Outcome 3 Exponentials and Logarithms

3. Logarithms ( Logs For Short ) REMINDER A function of the form f(x) = a x is called an EXPONENTIAL FUNCTION. The inverse function of the exponential function is called the LOGARITHMIC FUNCTION, and is written like this. f -1 ( x ) = log a x Also, a b = c  log a c = b

4. REMINDER OF HOW IT WORKS 2 3 = 8  log 2 8 = 3 4 2 = 16  log 4 16 = 2 5 4 = 625  log 5 625 = 4 log 2 64 = ?6 log 3 81 = ?4 Logarithms ( Logs For Short )

5. SHORT EXERCISE 1. log 3 27 = 2. log 9 81 = 3. log 2 128 = 4. log 9 3 = 5. log 8 2 = 6. log 3 243 = 7. log 2 ¼ = 8. log 3 = 1 81 3 2 7 0.5 1/31/3 5 -2 -4

6. The Three Rules For Logarithms There are three very important rules for logs which you must remember. They are... RULE 1log a (xy) = log a x + log a y RULE 2 RULE 3log a x n = nlog a x log a ( ) = log a x - log a y y x

7. PROOF OF RULE 1 Let log a x = pso that a p = x Let log a y = qso that a q = y So xy = a p a q = a p + q That is xy = a p + q EXPONENTIAL FORM So log a (xy) = p + q LOG FORM  log a (xy) = log a x + log a y

8. PROOF OF RULE 3 Let log a x = pSo a p = x Now let x n = (a p ) n = a pn So x n = a pn  log a x n = pn log a x n = np log a x n = nlog a x

9. EXAMPLES 1. Simplify log 12 72 + log 12 2 log 12 72 + log 12 2 = log 12 144 = 2 2. Simplify log 4 576 - log 4 9 log 4 576 - log 4 9 RULE 1 = log 4 64 = 3 RULE 2

10. 3. Simplify 2log 5 8 - 3log 5 4 2log 5 8 - 3log 5 4 = log 5 8 2 - log 5 4 3 = log 5 64 - log 5 64 = log 5 1 = 0 RULE 3 RULE 2 VERY IMPORTANT INDEED ! log1 in ANY base is ALWAYS 0

11. 4. Simplify 2log 9 2 + 3log 9 3 – log 9 36 2log 9 2 + 3log 9 3 – log 9 36 = log 9 2 2 + log 9 3 3 – log 9 36 = log 9 4 + log 9 27 – log 9 36 = log 9 108 - log 9 36 = log 9 3 = 0.5

12. 5. Solve the equation log 2 x + log 2 (x – 2) = 3 log 2 (x(x – 2)) = 3 log 2 (x 2 – 2x) = log 2 8 THE TRICK HERE IS TO CHANGE 3 INTO log 2 ( something ) x 2 – 2x = 8 x 2 – 2x – 8 = 0 (x – 4)(x + 2) = 0 x = 4 or x = -2

13. 6. Solve the equation log (x + 1) + log (x – 1) = log3 log ((x + 1)(x – 1)) = log3 log (x 2 – 1) = log3 (x 2 – 1) = 3 x 2 = 4 x = 2 or x = -2

14. Solving Exponential Equations EXAMPLE 1Solve the equation 4 x = 30 The trick is to take logs of both sides. So that you can use your calculator, use log to base 10. ( log 10 ) 4 x = 30 log 10 4 x = log 10 30 xlog 10 4 = log 10 30 log 10 30 log 10 4 x = x = 2.45 RULE 3

15. EXAMPLE 2 Solve the equation 20 x 7 x = 500 20 x 7 x = 500 7 x = 25 log 10 7 x = log 10 25 x log 10 7 = log 10 25 log 10 25 log 10 7 x = x = 1.65

16. Log To The Base e ; Problems There is a special log called log to the base e, where e is roughly equal to 2.71 ….. This is how they are written : log e and e x

17. EXAMPLE 1 The light intensity, I lumens, at a distance d metres away from the source, is given by the formula I(d) = I o e -0.03d. I o (the initial intensity of the light) is 80 lumens. Calculate: (a) The intensity of the light at a distance of 30 metres from the source. (b) The distance at which half the intensity is lost. (a) I(30) = 80e -0.03(30) I(d) = I o e -0.03d I(30) = 80e -0.9 I(30) = 32.5 lumens I(d) = 80e -0.03d

18. (b)If half the intensity is lost, then I (d) = 40 I(d) = 80e -0.03d 40 = 80e -0.03d 0.5 = e -0.03d log e 0.5 = log e e -0.03d TAKE LOGS log e 0.5 = -0.03d log e e Now log e e = 1 log e 0.5 = -0.03d d = 23.1 metres log e 0.5 -0.03 d = log e is written like this: ln

19. EXAMPLE 2 A radioactive mineral decays according to the formula m = m o e -0.05t, where m is the mass at time t years and m o is the initial mass in grams. Calculate : (a) The mass of the substance after 50 years, given that m o = 800. (b) The half-life of the substance. (a) m = m o e -0.05t m = 800e -0.05 x 50 m = 800e -2.5 m = 65.7 g m = 800e -0.05t

20. (b) Half life means that the mass is now 400g m = 800e -0.05t 400 = 800 e -0.05t 0.5 = e -0.05t log e 0.5 = log e e -0.05t log e 0.5 = -0.05tlog e e log e 0.5 = -0.05t log e 0.5 -0.05 t = t = 13.9 years

21. Relationships of the form y = kx m and y = ab x LOOK at this table of values x y 1 4 9 16 4.02.01.31.0 The graph looks like this: x y X X X X log 10 x log 10 y 0 0.60 0.95 1.20 0.60 0.30 0.11 0 Now look at this table

22. The graph now looks like this: log 10 x log 10 y X X X X log 10 x log 10 y 0 0.60 0.95 1.20 0.60 0.30 0.11 0

23. By taking logs of the values of x and y, we see that we now have a straight line. This means that the formula connecting log 10 x and log 10 y can be written in the same form as the equation of a straight line i.e. y = mx + c log 10 y = m log 10 x + c It would be better if we could find a formula which connects x and y rather than their logs. So ….

24. log 10 y = m log 10 x + c log 10 y = log 10 x m + c RULE 3 log 10 y = log 10 x m + log 10 k log 10 y = log 10 kx m RULE 1 y = kx m It is possible then to find a formula connecting x and y.

25. x y 1 4 9 16 4.0 2.0 1.3 1.0 EXAMPLE 1 For the table below, find a rule connecting x and y of the form y = kx m log 10 y = mlog 10 x + c log 10 y = log 10 x m + c log 10 y = log 10 x m + log 10 k log 10 y = log 10 kx m y = kx m

26. log 10 x log 10 y 0 0.60 0.95 1.20 0.60 0.30 0.11 0 log 10 y = m log 10 x + c To find m and c, choose two (well – spaced) points. 0 = -0.5 ( 1.20 ) + c c = 0.6 m = 0 - 0.60 1.20 - 0 = -0.5

27. log 10 y = -0.5 log 10 x + 0.6 THINK! I want to write 0.6 as log 10 (something) log 10 (something) = 0.6 (something) = 10 0.6 (something) = 4.0 log 10 y = -0.5 log 10 x + log 10 4.0 log 10 y = log 10 x -0.5 + log 10 4.0 log 10 y = log 10 ( 4.0x -0.5 ) y = 4.0x -0.5

28. EXAMPLE 2 log 10 V log 10 b 5 ( 12, 29 ) Find a formula connecting V and b. We have a straight line, so … log 10 V = m log 10 b + c From the graph, c = 5 m = 29 - 5 0.60 12 - 0 = 2

29. log 10 V = 2 log 10 b + 5 log 10 V = log 10 b 2 + 5 log 10 V = log 10 b 2 + log 10 100000 log 10 V = log 10 ( 100000b 2 ) V = 100000b 2 log 10 V = m log 10 b + c

30. A Slight Variation Sometimes, instead of being asked to find a rule in the form y = kx m, you might be asked to find a rule which looks like this: y = ab x y = ab x log 10 y = log 10 (ab x ) log 10 y = log 10 a + log 10 b x log 10 y = log 10 a + xlog 10 b log 10 y = xlog 10 b + log 10 a

31. EXAMPLE 3 x y 1 2 3 4 5.6 22.4 89.6 358.4 For the table below, find a rule connecting x and y of the form y = ab x. y = ab x log 10 y = log 10 (ab x ) log 10 y = log 10 a + log 10 b x log 10 y = log 10 a + xlog 10 b log 10 y = xlog 10 b + log 10 a

32. x log 10 y 1 2 3 4 0.75 1.35 1.95 2.55 log 10 y = xlog 10 b + log 10 a m = 2.55 - 0.75 0.60 4 - 1 = 0.6 0.75 = 0.6 (1) + c c = 0.15

33. a = 1.4 b = 10 0.6 y = 1.4(4.0) x y = ab x log 10 b = 0.6 b = 4.0 a = 10 0.15 log 10 a = 0.15

34. If y = kx m, the relationship can be expressed as a straight line if we change both axes to logarithmic ones. If y = ab x, we only change the y axis to a logarithmic one to express the relationship as a straight line. Important