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

2. 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 = ½ Ex.3 Question 1

3. Solve log a (x + 1) + log a (x – 1) = log a 3 log a (x + 1) + log a (x – 1) = log a 3 log a ((x + 1)(x – 1)) = log a 3 log a (x 2 – 1) = log a 3 (x 2 – 1) = 3 x 2 = 4 x = 2 or x = -2 Ex.3 Question 2(a)

4. Solve log 2 x + log 2 (x – 2) = 3 log 2 x + log 2 (x – 2) = 3 log 2 (x(x – 2)) = 3 log 2 (x 2 – 2x) = 3 x 2 – 2x = 8 x 2 – 2x – 8 = 0 (x – 4)(x + 2) = 0 x = 4 or x = -2 x 2 – 2x = 2³ Ex.3 Question 2(b)

5. Solve 4 x = 30 The trick is to take logs of both sides. So that you can use your calculator, use log to the base 10 (log 10 ). 4 x = 30 log 10 4 x = log 10 30 xlog 10 4 = log 10 30 x = 2.45 log 10 30 log 10 4 x = Ex.3 Question 3(a)

6. Solve 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 x = 1.65 log 10 25 log 10 7 x = Ex.3 Question 3(b)

7. 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 Ex.3 Question 4

8. 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 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 = (b) Ex.3 Question 4 cont.

9. 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. m = m o e -0.05t m = 800e -0.05 x 50 m = 800e -2.5 m = 65.7 g m = 800e -0.05t Ex.3 Question 5 (a)

10. 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 (b) Ex.3 Question 5 cont.

11. As shown in the diagram below, a set of experimental results give a straight line when log 10 V is plotted against log 10 b. Express V in terms of b. Ex.3 Question 6 log 10 V log 10 b 5 (12,29)We have a straight line, so… log 10 V = m log 10 b + c From the graph, c = 5 m = 29 - 5 12 - 0 = 2 log 10 V = m log 10 b + c log 10 V = 2 log 10 b + 5 5

12. As shown in the diagram below, a set of experimental results give a straight line when log 10 V is plotted against log 10 b. Express V in terms of b. Ex.3 Question 6 cont. log 10 V log 10 b 5 (12,29) log 10 V = 2 log 10 b + 5 log 10 V = log 10 b 2 + log 10 100000 log 10 V = log 10 (100000b 2 ) V = 100000b 2 log 10 V = log 10 b 2 + 5