PSf Eponentials and Logs Paper 1 Section A Each correct answer in this section is worth two marks. 1. Simplif log 4 8 + log 4 2 3 log 5 5. A. 1 2 B. 1 C. log 4 ( 165 ) ( ) D. log 16 4 125 Ke utcome Grade Facilit Disc. Calculator Content Source B 3.3 A/B 0.53 0.33 NC A28 HSN 17 PSf hsn.uk.net Page 1
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 0.64 0.59 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 0.7 0.45 CN A28, A32 HSN 111 PSf hsn.uk.net Page 2
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. 1 18 log e 4 3 Ke utcome Grade Facilit Disc. Calculator Content Source B 3.3 A/B 0.43 0.31 NC A30 HSN 095 PSf hsn.uk.net Page 3
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 0.56 0.77 CN A31 HSN 112 PSf Paper 1 Section B [END F PAPER 1 SECTIN A] 6. Evaluate log 5 2 + log 5 50 log 5 4. 3 Part Marks Level Calc. Content Answer U3 C3 2 C NC A28 2 2000 P1 Q9 1 A/B NC A28 1 2 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 5 100 log 5 4 2 log 5 25 3 2 7. Evaluate log 4 16 + 2 log 3 (3 3). 3 Part Marks Level Calc. Content Answer U3 C3 3 A NC A28 5 B 11-004 1 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 4 16 + 2 log 3 (3 3) = 5
PSf 8. Evaluate log 3 18 + log 3 6 log 3 4 + log 5 5. 4 Part Marks Level Calc. Content Answer U3 C3 2 B CN A28 E 3-3-2 7 2 C CN A28 2 1 2 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 ( 18 6 4 ) 3 log 3 3 3 + log 5 5 1 2 = 3 log 3 3 + 1 2 log 5 5 4 3 + 1 2 = 7 2 9. 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 3-3-4 1 2 pd: use log laws pd: process 3 ic: state n 1 4(log a 2 + log a a) 3 log a 2 2 4 + 4 log a 2 3 log a 2 = 4 + log a 2 3 n = 2 hsn.uk.net Page 5
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 11-003 1 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
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
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 = 1 4 5 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 2 + 1 6 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
PSf 14. Given = log 5 3 + log 5 4, find algebraicall the value of. 4 15. 16. Find if 4 log 6 2 log 4 = 1. 3 Part Marks Level Calc. Content Answer U3 C3 3 C NC A32, A28, A31 = 81 2001 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 4 2 2 log 6 4 4 2 3 all processing leading to = 81
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 2 2 2 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 3-3-6 1 ss: use laws of logs 2 pd: use laws of logs 3 pd: use laws of logs 1 log 8 = log 6 9 + log 6 4 = log 6 36 2 log 8 = 2 3 = 8 2 = 64 19. 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 = 5 2 2002 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 ) + 1 2 log 5 = log 5 2 +... 3... + log 5 5 4 = 5 2 hsn.uk.net Page 10
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 11-002 1 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 4 + 3 3 log 4 = log 4 k + n log 4 4 log 4 k = 3 k = 4 3 5 = 64 [END F PAPER 1 SECTIN B] hsn.uk.net Page 11
PSf Paper 2 1. Solve log 2 2 3 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 11-001 1 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 = 7 16 1 hsn.uk.net Page 12
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 = 2 2 3 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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PSf 5. 6. 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 = 0 46 2001 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 1 5 2 e.g. 1 5k = ln 2 3 k = 0 46 hsn.uk.net Page 15
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 0 01225t. (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 3-3-3 (b) 3 B CR A30 l = 81 633 (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 2 3 4 N(l) = en 0 = 32e 32e 0 01225l = 32e 0 01225l = 1 l = 81 633 (3 d.p.) 8. hsn.uk.net Page 16
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PSf 15. 16. 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 A31 2002 P2 Q7 1 A/B CN A32 = 2 1 3 1 ss: know to isolate log term 2 pd: epress log equation as ep. equ. 3 pd: process 1 2 log 3 ( 2) = 1 2 = 3 1 3 = 2 1 3 hsn.uk.net Page 22
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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 3 54. 4 (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, A28 4 82 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 3 4 + 3 2 = log 3 (4 1/2 ) + 3 3 = log 3 2 + log 3 3 3 4 = log 3 54 5 u 2 = 1 2 log 3 54 + 3 6 = log 3 (54 1/2 ) + log 3 3 3 7 = log 3 (27 54) (= 81 6) 8 3 u 2 = 81 6. 9 u 2 log e 3 = log e 81 6 10 u 2 = (log e 81 6)/(log e 3) 11 = 4 82 to 2 d.p. 20. hsn.uk.net Page 24
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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 + 1 8 2000 P2 Q11 (b) 4 A/B CR A33 a = 6 05, b = 0 6 1 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 = 1 8 3 = 0 6 2 P = 0 6Q + 1 8 Method 1 3 log e p = 0 6 log e q + 1 8 4 log e q 0 6 5 log e 6 05 6 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 + 1 8 stated or implied b 5 or 6 5 ln a = 1 8 6 a = 6 05, b = 0 6 3 hsn.uk.net Page 26
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PSf 24. The results of an eperiment were noted as follows. 1 70 2 10 2 50 2 90 log 10 2 14 1 96 1 75 1 53 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 = 1023 29 (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 4 1 53 2 14 m = = 0 51 (2 d.p.), 2 90 1 70 so b = 10 0 51 = 0 31 (2 d.p.) 5 (1 70, 2 14) : 2 14 = 0 51 1 70 + log 10 a 6 a = 1023 29 (2 d.p.) so = 1023 29 (0 31) 25. hsn.uk.net Page 28
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PSf 27. [END F PAPER 2] hsn.uk.net Page 30