Exponentials and Logs

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1 PSf Eponentials and Logs Paper 1 Section A Each correct answer in this section is worth two marks. 1. Simplif log log log 5 5. A. 1 2 B. 1 C. log 4 ( 165 ) ( ) D. log Ke utcome Grade Facilit Disc. Calculator Content Source B 3.3 A/B NC A28 HSN 17 PSf hsn.uk.net Page 1

2 PSf 2. Solve log b log b 7 = log b 3 for > 0. A. = 21 B. = 10 C. = 3 7 D. = 3 7 Ke utcome Grade Facilit Disc. Calculator Content Source A 3.3 A/B CN A28, A32 HSN 175 PSf 3. Solve log a 5 + log a = log a 20 for > 0. A. = 1 4 B. = 4 C. = 15 D. = 100 Ke utcome Grade Facilit Disc. Calculator Content Source B 3.3 A/B CN A28, A32 HSN 111 PSf hsn.uk.net Page 2

3 PSf 4. The diagram shows the graph of = 3e k. PSf 3 = 3e k (4, 18) What is the value of k? 3 A. 2e B. 1 4 log e 6 C. 1 4 log e 15 D log e 4 3 Ke utcome Grade Facilit Disc. Calculator Content Source B 3.3 A/B NC A30 HSN 095 PSf hsn.uk.net Page 3

4 PSf 5. Solve 3 log a 2 = 1 2 A. a = 64 for a. B. a = 36 C. a = 4 9 D. a = 1 16 Ke utcome Grade Facilit Disc. Calculator Content Source A 3.3 A/B CN A31 HSN 112 PSf Paper 1 Section B [END F PAPER 1 SECTIN A] 6. Evaluate log log 5 50 log Part Marks Level Calc. Content Answer U3 C3 2 C NC A P1 Q9 1 A/B NC A pd: use log a + log a = log a pd: use log a log a = log a 3 pd: use log a a = 1 1 log log log Evaluate log log 3 (3 3). 3 Part Marks Level Calc. Content Answer U3 C3 3 A NC A28 5 B ss: use laws of logs 2 ss: use laws of logs 3 pd: complete hsn.uk.net Page 4 1 log 4 16 = 2 2 log 3 (9 3) 3 log log 3 (3 3) = 5

5 PSf 8. Evaluate log log 3 6 log log Part Marks Level Calc. Content Answer U3 C3 2 B CN A28 E C CN A pd: use log a + log a = log a pd: use log a log a = log a 3 pd: use log a k = k log a 4 pd: use log a a = 1 1 log 3 (18 6) 2 log 3 ( ) 3 log log = 3 log log = The epression 4 log a (2a) 3 log a 2 can be written in the form 2n + log a n, where n is a whole number. Find the value of n. 3 Part Marks Level Calc. Content Answer U3 C3 3 B CN A28 n = 2 E pd: use log laws pd: process 3 ic: state n 1 4(log a 2 + log a a) 3 log a log a 2 3 log a 2 = 4 + log a 2 3 n = 2 hsn.uk.net Page 5

6 PSf 10. The graph below shows the curve with equation = log 8. PSf (8, 1) = log 8 1 Sketch the graph of = log 8 ( 1 2 ). 3 Part Marks Level Calc. Content Answer U3 C3 3 A CN A28, A3 sketch B ss: use laws of logs = 2 log 8 2 ic: know to reflect and scale PSf 2 reflect in -ais and scale 3 ic: annotate sketch 3 show (1, 0) and (8, 2) 1 = log 8 ( 1 2 ) 1 (8, 2) 11. The diagram shows the curve with equation = log 2. PSf = log 2 1 Sketch the curve = log 2 ( 2 ). 4 Part Marks Level Calc. Content Answer U3 C3 4 A CN A29, A28 sketch AT064 1 ss: use law of logs 2 ic: interpret reflection 3 ic: interpret translation 4 ic: annotate sketch hsn.uk.net Page 6 = log 2 2 log 2 reflect curve in -ais shift 1 unit up 4 decreasing log curve thro (1, 1) 1 2 3

7 PSf 12. Sketch and annotate the curve with equation = log e ( + 1) 2. 4 Part Marks Level Calc. Content Answer U3 C3 4 A CN A29, A28 sketch WCHS U3 Q10 1 ss: use log law 2 ic: interpret translation 3 ic: interpret scaling 4 ic: sketch with points annotated 1 = 2 log e ( + 1) 2 = log e translated 1 unit left 3 then made twice as tall 4 sketch, with points (0, 0) and (e 1, 2) hsn.uk.net Page 7

8 PSf 13. The diagram below shows the graph of the function f () = e. PSf Q = f () P The points P and Q have -coordinates 1 and 1 respectivel. Straight lines are drawn from the origin to P and Q as shown. (a) Show that P and Q are perpendicular. 3 (b) Show that the area of triangle PQ is 1 + e2 2e (c) The function g is defined b g() = f ( 2) + 1. (i) Sketch the curve with equation = g().. 3 (ii) The graphs of = f () and = g() intersect when = a. Show that e a = e2 e 2 1. Hence epress a in the form A + B log e (e 1) + C log e (e + 1), stating the values of A, B and C. 9 Part Marks Level Calc. Content Answer U3 C3 (a) 3 B CN G5, G2 proof AT040 (b) 3 B CN G1 proof (ci) 3 C CN A3 sketch (cii) 6 A CN A30, A28 A = 2, B = C = 1 1 ic: obtain coordinates of P and Q 2 pd: find gradients 3 ss: use m m = pd: compute side length pd: compute side length 6 ss: use area formula and complete 7 ss: translate parallel to -ais 8 ss: translate parallel to -ais 9 ic: annotations 10 ss: form equation 11 ic: complete 12 ss: know to take logs hsn.uk.net 13 pd: use laws of logs Page 8 14 pd: use laws of logs 15 ic: state A, B, C 1 P ( 1, 1 ) e, Q(1, e) 2 m P = 1 e, m Q = e 3 m P m Q = 1 so P Q 4 P = 1 + 1/e 2 5 Q = e Area = 1 2 P Q = (1 + e2 )/2e shift 2 units to right shift 1 unit up 9 P (1, 1/e + 1), Q (3, e + 1) 10 f () = g() 11 e = e 2 /(e 2 1) 12 ( = logquestions e e 2 /(e 2 marked 1) ) c SQA 13 = log All e e 2 others log e (e c 2 Higher 1) Still Notes 14 = 2 log e (e + 1) log e (e 1) 15 A = 2, B = C = 1 7 8

9 PSf 14. Given = log log 5 4, find algebraicall the value of Find if 4 log 6 2 log 4 = 1. 3 Part Marks Level Calc. Content Answer U3 C3 3 C NC A32, A28, A31 = P1 Q8 1 2 pd: use log-to-inde rule pd: use log-to-division rule 3 ic: interpret base for log a = 1 and simplif hsn.uk.net Page 9 1 log 6 4 log log all processing leading to = 81

10 PSf 17. Solve log a 12 + log a 2 log a 2 = 6 for in terms of a, where a > 1. 4 Part Marks Level Calc. Content Answer U3 C3 4 B CN A32, A28, A31 a = 1 3 a6 WCHS U3 Q6 1 ss: use log law 2 ss: use log law 3 ss: know to convert log to eponential 4 pd: complete 1 log a 12 + log a log a log a ( 12 4 ) 3 3 = a 6 4 a = 1 3 a6 18. Solve log 8 log 6 4 = log 6 9 for > 0. 3 Part Marks Level Calc. Content Answer U3 C3 3 C CN A32, A28, A31 = 64 E ss: use laws of logs 2 pd: use laws of logs 3 pd: use laws of logs 1 log 8 = log log 6 4 = log log 8 = 2 3 = 8 2 = The graph illustrates the law PSfrag = k n. log 5 If the straight line passes through A(0 5, 0) and B(0, 1), find the values of B(0, 1) k and n. 4 A(0 5, 0) log 5 Part Marks Level Calc. Content Answer U3 C3 4 A/B NC A33 = P1 Q11 1 ic: interpret graph 2 ss: use log laws 3 ss: use log laws 4 pd: solve log equation 1 log 5 = 2(log 5 ) log 5 = log log = 5 2 hsn.uk.net Page 10

11 PSf 20. The graph below shows a straight line in the (log 4, log 4 )-plane. log PSf 4 (7, 10) 3 log 4 Find the equation of the line in the form = k n, where k, n R. 5 Part Marks Level Calc. Content Answer U3 C3 5 A NC A33, G3, A28 = 64 B ic: interpret graph (gradient) 2 ic: interpret graph (complete eqn) 3 ss: use log laws 4 ss: use log laws 5 ic: complete 1 gradient = 1 2 log 4 = log log 4 = log 4 k + n log 4 4 log 4 k = 3 k = = 64 [END F PAPER 1 SECTIN B] hsn.uk.net Page 11

12 PSf Paper 2 1. Solve log log 2 = 4 log 2 7 for > 0. 3 Part Marks Level Calc. Content Answer U3 C3 3 A CN A28, A31 = 7 16 B ss: use log law 2 ss: use log law 3 ss: know to convert log to eponential and complete log 2 2 = 2 log 2 2 log 2 7 log 2 = log 2 ( 7 ) 3 7 = 24, so = hsn.uk.net Page 12

13 PSf 2. The graph of the function f () = log e is shown in the diagram below. PSf = f () B(e 2, b) A(1, 0) The point B has coordinates (e 2, b). (a) Write down the value of b. 1 (b) The function g is defined b g() = f ( 2). Sketch the graph of = g(). 3 (c) The graphs of = f () and = g() intersect at C. The -coordinate of C is of the form = m + n. Determine the values of m and n. 6 Part Marks Level Calc. Content Answer U3 C3 (a) 1 C CN A2 2 AT010 (b) 3 C CN A29 sketch (c) 6 A CN A32, A34 m = 1, n = 2 1 ic: interpret graph 2 ic: reflection 3 ic: horizontal translation 4 ic: annotate sketch pd: epression for g() 6 ss: equate 7 ss: use log law 8 ss: convert from log 9 pd: solve quadratic equation 10 ic: interpret solution 5 1 b = reflect in -ais shift 2 units to right 4 show A (3, 0) and B (e 2 + 2, 2) 5 g() = log 2 ( 2) 6 log e = log e ( 2) 7 log e = log e ( 2) 1 8 = 1/( 2) 9 = 1 ± 2 10 m = 1, n = 2 hsn.uk.net Page 13

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15 PSf Before a forest fire was brought under control, the spread of the fire was described b a law of the form A = A 0 e kt where A 0 is the area covered b the fire when it was first detected and A is the area covered b the fire t hours later. If it takes one and a half hours for the area of the forest fire to double, find the value of the constant k. 3 Part Marks Level Calc. Content Answer U3 C3 3 A/B CR A30 k = P2 Q9 1 ic: form eponential equation 2 ss: epress ep. equ. as log equation 3 pd: solve log equation 1 2A 0 = A 0 e k e.g. 1 5k = ln 2 3 k = 0 46 hsn.uk.net Page 15

16 PSf 7. A population of bacteria is growing in such a wa that the number of bacteria N present after t minutes is given b the formula N(t) = 32e t. (a) State N 0, the number of bacteria present when t = 0. 1 (b) The e-folding time, l minutes, is the length of time until N(l) = en 0. Find the e-folding time for this population correct to 3 decimal places. 3 Part Marks Level Calc. Content Answer U3 C3 (a) 1 C CN A6 N 0 = 32 E (b) 3 B CR A30 l = (3 d.p.) 1 ic: interpret formula 1 N 0 = N(0) = 32 2 ic: interpret N(l) 3 ss: form equation 4 pd: solve N(l) = en 0 = 32e 32e l = 32e l = 1 l = (3 d.p.) 8. hsn.uk.net Page 16

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22 PSf Find the -coordinate of the point where the graph of the curve with equation = log 3 ( 2) + 1 intersects the -ais. 3 Part Marks Level Calc. Content Answer U3 C3 2 C CN A P2 Q7 1 A/B CN A32 = ss: know to isolate log term 2 pd: epress log equation as ep. equ. 3 pd: process 1 2 log 3 ( 2) = 1 2 = = hsn.uk.net Page 22

23 PSf hsn.uk.net Page 23

24 PSf 19. A sequence is defined b the recurrence relation u n+1 = 2 1u n + 3 with u 0 = log 3 4. (a) Show that u 1 = log (b) Find an epression for u 2 in the form log 3 a. 3 (c) Find the value of u 2 correct to two decimal places. 4 Part Marks Level Calc. Content Answer U3 C3 (a) 4 A CN A11, A28 proof AT051 (b) 3 A CN A11, A28 log 3 (81 6) (c) 4 A CR A31, A to 2 d.p. 1 ic: find u 1 2 ss: use law of logs 3 ss: convert constant to log 4 ss: use law of logs 5 ic: find u 2 6 ss: use law of logs 7 pd: complete 8 ss: know to change base 9 ss: use law of logs 10 pd: process 11 ic: state value 1 u 1 = 1 2 log = log 3 (4 1/2 ) = log log = log u 2 = 1 2 log = log 3 (54 1/2 ) + log = log 3 (27 54) (= 81 6) 8 3 u 2 = u 2 log e 3 = log e u 2 = (log e 81 6)/(log e 3) 11 = 4 82 to 2 d.p. 20. hsn.uk.net Page 24

25 PSf 21. hsn.uk.net Page 25

26 Higher Mathematics 22. PSf The results of an eperiment give rise to the graph shown. PSf (a) Write down the equation of the line in 1 8 terms of P and Q. 2 3 Q P It is given that P = log e p and Q = log e q. (b) Show that p and q satisf a relationship of the form p = aq b, stating the values of a and b. 4 Part Marks Level Calc. Content Answer U3 C3 (a) 2 A/B CR G3 P = 0 6Q P2 Q11 (b) 4 A/B CR A33 a = 6 05, b = ic: interpret gradient 2 ic: state equ. of line 3 ic: interpret straight line 4 ss: know how to deal with of log 5 ss: know how to epress number as log 6 ic: interpret sum of two logs 1 m = = P = 0 6Q Method 1 3 log e p = 0 6 log e q log e q log e p = 6 05q 0 6 Method 2 ln p = ln aq b ln p = ln a + b ln q 4 ln p = 0 6 ln q stated or implied b 5 or 6 5 ln a = a = 6 05, b = hsn.uk.net Page 26

27 PSf 23. hsn.uk.net Page 27

28 PSf 24. The results of an eperiment were noted as follows log The relationship between these data can be written in the form = ab where a and b are constants. Find the values of a and b and hence state a formula relating the data. 6 Part Marks Level Calc. Content Answer U3 C3 6 A CR A33, A28 = (0 31) WCHS U3 Q13 1 ss: know to take logs 2 pd: use laws of logs 3 ic: interpret equation 4 pd: find gradient 5 pd: start to find other constant 6 ic: complete, and state equation 1 log 10 = log 10 (a ) 2 log 10 = (log 10 b) + log 10 a 3 gradient of line is log 10 b m = = 0 51 (2 d.p.), so b = = 0 31 (2 d.p.) 5 (1 70, 2 14) : 2 14 = log 10 a 6 a = (2 d.p.) so = (0 31) 25. hsn.uk.net Page 28

29 PSf 26. hsn.uk.net Page 29

30 PSf 27. [END F PAPER 2] hsn.uk.net Page 30

Polynomials and Quadratics

Polynomials and Quadratics PSf Paper 1 Section A Polnomials and Quadratics Each correct answer in this section is worth two marks. 1. A parabola has equation = 2 2 + 4 + 5. Which of the following are true? I. The parabola has a