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Scholar Higher Mathematics Homework Session Thursday 19 th March 7:30pm You will need a pencil, paper and a calculator for some of the activities.

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Presentation on theme: "Scholar Higher Mathematics Homework Session Thursday 19 th March 7:30pm You will need a pencil, paper and a calculator for some of the activities."— Presentation transcript:

1 Scholar Higher Mathematics Homework Session Thursday 19 th March 7:30pm You will need a pencil, paper and a calculator for some of the activities

2 SCHOLAR online tutor for Maths and Author of the new SCHOLAR National 5 & Higher Maths courses Margaret Ferguson

3 Logs & Exponentials and Recurrence Relations Tonight’s Revision Session will cover

4 Logs and exponentials are inverses of each other If y = a x then log a y = x If log a x = y then a y = x Exponential functions take the form y = a x a > 1 for a growth function 0 < a < 1 for a decay function The exponential function is e x where e is … Logs & Exponentials things you should know e.g. 7 = 5 x => log 5 7 = x e.g. log 3 x = 2 then 3 2 = x

5 Logs & Exponentials more things you should know Laws of exponentials a m x a n = a m + n (a m ) n = a mn a 0 = 1 log a m + log a n = log a (mn) Laws of logarithms log a x n = n log a x log a 1 = 0 log a a = 1

6 Which of the following sketches shows the graph of y = log 6 2 x ? Vote for the correct answer now x y x y x y x y (6,1) (3,1) (1,0) (½,0) (a) (b) (c) (d) The graph crosses the x -axis when y = 0 log 6 2 x = 0 => 2x = 1 so x = ½ and (½,0) lies on the graph The other point on the graph occurs when y = 1 log 6 2 x = 1 => 2x = 6 so x = 3 ✓ 6 0 = 2 x 6 1 = 2 x and the point (3,1) lies on the graph

7 A = 2 x A 0 and t = 1.5 2A 0 = A 0 e 1.5k 2 = e 1.5k log e 2 = 1.5k k = log e = ln A = A 0 e kt

8 ✓ A B C D 2 Given that log log 4 q = 1, what is the value of q? Vote for the correct answer now log log 4 q = 1 log 4 8q = = 8q

9 Experimental data of the formy = kx n apply logs to both sides of the equation and choose a base log a y = log a kx n log a y = log a k + log a x n log a y = log a k + n log a x log a y = n log a x + log a k This equation looks a bit like y = mx + c but is written as Y = mX + c where where Y = log a y, m = n, X = log a x and c = log a k To find the equation from the graph of log a x against log a y n = m where m is the gradient k = a c where c is the value of the y-intercept

10 The graph illustrates the law y = kx n log 5 x log 5 y B(0,1) A(0.5,0) The straight line passes through the points A(0.5,0) and B(0,1). What are the values of k and n? Step 1: Find the equation of the line. and B(0,1) so y – 1 = -2(x – 0) y = -2x + 1 log 5 y = -2log 5 x + 1 Remember y = kx n gives log 5 y = log 5 kx n log 5 y = log 5 k + log 5 x n log 5 y = log 5 k + n log 5 x log 5 y = n log 5 x + log 5 k Hence n = -2 and log 5 k = 1=>5 1 = k giving the equation y = 5x -2 Step 2: Identify k and n. log 5 y = n log 5 x + log 5 k

11 Experimental data of the formy = ab x apply logs to both sides of the equation and choose a base log 2 y = log 2 ab x log 2 y = log 2 a + log 2 b x log 2 y = log 2 a + x log 2 b log 2 y = x log 2 b + log 2 a This equation looks a bit like y = mx + c but is written as Y = mx + c where Y = log 2 y, m = log 2 b and c = log 2 a To find the equation from the graph of x against log 2 y b = 2 m where m is the gradient a = 2 c where c is the value of the y-intercept

12 The results of an experiment give rise to the graph shown below. It is given that P = log e p and Q = q. Show that p and q satisfy a relationship of the form p = ab q. If p = ab q then log e p = log e ab q log e p = log e a + log e b q log e p = log e a + qlog e b log e p = qlog e b + log e a P = 0.6Q log e p = 0.6q Hence log e b = 0.6 and log e a = 1.8 e 0.6 = b and e 1.8 = a b = 1.82 and a = 6.05 giving p = 6.05 x 1.82 q log e p = qlog e b + log e a

13 ✓ A x = 10 x 3 y B x = 30 10y C x = 3y + 10 D x = y Given that log 10 (x) = y log 10 (3) + 1, express x in terms of y. Vote for the correct answer now log 10 x = y log log 10 x = y log log log 10 x = log 10 3 y + log log 10 x = log 10 (3 y x 10) x = 3 y x 10

14 A sequence is a pattern of numbers that can be defined by a rule or formula. A recurrence relation describes a sequence where each term is a function of the previous term. A geometric sequence takes the form u n + 1 = au n An arithmetic sequence takes the form u n + 1 = u n + b A linear recurrence relation can defined by u n + 1 = au n + b (a ≠ 0) For a linear recurrence relation a limit exists if -1 < a < 1 The limit is given by the formula Recurrence Relations

15 A sequence is defined by the recurrence relation u n+1 = 0.6u n + k and u 0 = 3. As n -> ∞, the limit of this sequence is 5. What is the value of k ? A0 B0.88 C2 D8 Vote for the correct answer now ✓ k = 5 x 0.4 = 2

16 A recurrence relation is defined by u n + 1 = pu n + q, where -1 < p < 1 and u 0 = 12. If u 1 = 15 and u 2 = 16, find the values of p and q. Hence or otherwise find the limit of this recurrence relation. What is an expression for u 1 ? What is an expression for u 2 ? 15 = 12p + q 16 = 15p + q subtract 1 = 3p p = ⅓ 15 = 12 x ⅓ + q 15 = 4 + q q = 11 u n + 1 = ⅓u n + 11 a limit exists because -1 < ⅓ < 1

17 The floors of a shopping centre are cleaned daily. Grimex removes 70% of all germs but during the next 24 hours, 300 “new” germs per sq unit are estimated to appear. Carefree removes 80% of all germs but during the next 24 hours, 350 “new” germs per sq unit are estimated to appear. For Grimex let u n represent the number of germs per unit 2 on the floor immediately before disinfecting for the n th time and v n for Carefree. (a)Write down a recurrence relation for each product prior to disinfecting. (b)Determine which product is more effective in the long term. Grimex is more effective in the long term as the number of germs will settle around 428 germs per unit 2 which is less that Carefree which will settle around 437 germs per unit 2. and v n+1 = 0.2v n u n+1 = 0.3u n + 300

18 Question Time If you have any questions about tonight’s session please ask The next session will be on 30 th April at 7:30pm The session will cover revision for the Exam Please give us feedback on tonight’s session You will find a link in the chat box


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