IB Maths SL Logarithm Practice Problems Mr. W Name

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1 IB Maths SL Logarithm Practice Problems Mr. W Name This packet is made up of three parts. The first deals with the basics of logarithms, their representations and their graphs. The second deals with using properties and representations of logarithms to solve logarithmic equations, and the third uses logarithms to solve eponential equations some of which you already know how to solve a different way. Each part contains its own mark scheme. Please do on separate paper as you will be handing some of them in for credit. This may require you to copy graphs on separate paper, as well. Part I: Representations, Rules and Graphs of Logarithms. Let f() = log + log 6 log 4, for > 0. (a) Show that f() = log. () (b) Find the value of f(0.5) and of f(4.5). () ln a The function f can also be written in the form f() =. ln b (c) (i) Write down the value of a and of b. (ii) Hence on graph paper, sketch the graph of f, for 5 5, 5 y 5, using a scale of cm to unit on each ais (IF YOU DON T HAVE CM PAPER, LET BOX = CM) (iii) Write down the equation of the asymptote. (6) (d) Write down the value of f (0).() The point A lies on the graph of f. At A, = 4.5. (e) On your diagram, sketch the graph of f, noting clearly the image of point A. (4). Let f() = log, for > 0. (a) Show that f () =. (b) Write down the range of f. () () Let g() = log, for > 0. (c) Find the value of (f g)(), giving your answer as an integer. (4) (Total 7 marks)

2 . Let f() = ln and g() = ln 5. (a) Epress g() in the form f() + ln a, where a +. (4) (b) The graph of g is a transformation of the graph of f. Give a full geometric description of this transformation. () (Total 7 marks) 4. (a) Find log. (b) Given that log can be written as p + qy, find the value of p and of q. y 8 () (4) (Total 5 marks) 5. Let f() = k log. (a) Given that f () = 8, find the value of k. (b) Find f. () (4) (Total 7 marks) 6. Given that p = log a 5, q = log a, epress the following in terms of p and/or q. (a) log a 0 (b) log a 8 (c) log a.5 7. (a) Let log c = p and log c 5 = q. Find an epression in terms of p and q for (i) log c 5; (ii) log c 5. (b) Find the value of d if log d 6 =.

3 8. Let ln a = p, ln b = q. Write the following epressions in terms of p and q. (a) ln a b (b) ln a b 9. The diagram shows three graphs. y B A C A is part of the graph of y =. B is part of the graph of y =. C is the reflection of graph B in line A. Write down (a) (b) the equation of C in the form y =f (); the coordinates of the point where C cuts the -ais. Answers: (a)... (b)... (Total 4 marks)

4 0. Let p = log 0, q = log 0 y and r = log 0 z. Write the epression log 0 in terms of p, q and r. y z Answer:... Find the eact value of in each of the following equations. (a) 5 + = 65 (b) log a ( + 5) =. Solve the following equations. (a) ln ( + ) =. (b) 0 = Let f () = log a, > 0. (a) Write down the value of (i) f (a); (ii) f (); (iii) f (a 4 ). ()

5 (b) The diagram below shows part of the graph of f. y f 0 On the same diagram, sketch the graph of f. () 4. Let a = log, b = log y, and c = log z. Write log y in terms of a, b and c. z Answer: Let f () = ln ( + ), > and g () = e ( 4), > 0. (a) Write down the -intercept of the graph of f. (b) (i) Write down f (.999). (ii) Find the range of f.

6 (c) Find the coordinates of the point of intersection of the graphs of f and g. 6. Given that log 5 = y, epress each of the following in terms of y. (a) log 5 (b) log 5 (c) log 5 Answers: (a)... (b)... (c) The functions f () and g () are defined by f () = e and g () = ln (+ ). (a) Write down f (). (b) (i) Find ( f g) (). (ii) Find ( f g) ().

7 P 8. Let log 0 P =, log 0 Q = y and log 0 R = z. Epress log 0 in terms of, y and z. QR Answer:... (Total 4 marks) 9. If log a = and log a 5 = y, find in terms of and y, epressions for (a) log 5; (b) log a 0. Answers: (a)... (b)... (Total 4 marks)

8 Markscheme for Part I:. (a) combining terms e.g. log 8 log 4, log + log 4 epression which clearly leads to answer given A e.g. 8 4 log, log f() = log AG N0 (b) attempt to substitute either value into f (M) e.g. log, log 9 f(0.5) = 0, f(4.5) = AA N (c) (i) a =, b = AA NN (ii) Note: Award A for sketch approimately through (0.5 ± 0., 0 ± 0.) AAA N

9 A for approimately correct shape, A for sketch asymptotic to the y-ais. (iii) = 0 (must be an equation) A N (d) f (0) = 0.5 A N (e) Note: Award A for sketch approimately through (0 ± 0., 0.5 ± 0.), A for approimately correct shape of the graph reflected over y =, A for sketch asymptotic to -ais, A for point ( ± 0., 4.5 ± 0.) clearly marked and on curve. AAAA 4 N4. (a) interchanging and y (seen anywhere) (M) e.g. = log y (accept any base) evidence of correct manipulation e.g. y = y, =, = log y, y = log f () = AG N0 A

10 (b) y > 0, f () > 0 A N (c) METHOD finding g() = log (seen anywhere) attempt to substitute e.g. (f g)() = log A (M) evidence of using log or inde rule e.g. (f g)() = log 4 log, (f g)() = 4 A N METHOD attempt to form composite (in any order) e.g. (f g)() = log (M) evidence of using log or inde rule e.g.(f g)() = log log, (f g)() = A (f g)() = 4 A N [7]. (a) attempt to apply rules of logarithms (M) e.g. ln a b = bln a, ln ab = ln a + ln b correct application of ln a b = bln a (seen anywhere) A e.g. ln = ln correct application of ln ab = ln a + ln b (seen anywhere) A e.g. ln 5 = ln 5 + ln so ln 5 = ln 5 + ln g () = f () + ln5 (accept g () = ln + ln 5) A N 4 (b) transformation with correct name, direction, and value A 0 e.g. translation by, shift up by ln 5, vertical translation of ln 5 ln5 [7]

11 4. (a) 5 A N (b) METHOD y log log log 8 y 8 = = log y log 8 log 8 = p = 5, q = (accept 5 y) A N METHOD 8 y 5 ( ) = ( ) y 5 = y = 5 y log ( 5 y ) = 5 y p = 5, q = (accept 5 y) A N [5] 5. (a) METHOD recognizing that f(8) = e.g. = k log 8 recognizing that log 8 = e.g. = k (M) k = A N METHOD attempt to find the inverse of f() = k log e.g. = k log y, y = substituting and 8 k k e.g. = k log 8, = 8 (M) (M) k = k = A N log 8

12 (b) METHOD recognizing that f() = (M) e.g. = log log = f = 4 (accept = 4) A N METHOD attempt to find inverse of f() = log (M) k e.g. interchanging and y, substituting k = into y = correct inverse e.g. f () =, f = 4 A N [7] 6. (a) log a 0 = log a (5 ) (M) = log a 5 + log a = p + q A N (b) log a 8 = log a (M) = log a = q A N (c) 5 log a.5 = log a (M) = log a 5 log a = p q A N 7. (a) (i) log c 5 = log c + log c 5 = p + q A N (ii) log c 5 = log c 5 = q A N

13 (b) METHOD d = 6 M d = 6 A N METHOD For changing base M eg log log = d, log 0 6= log 0 d d = 6 A N 8. (a) ln a b = ln a + ln b ln a b = p + q A N a (b) ln = ln a ln b b a ln = p q A N b 9. (a) C has equation = y ie y = log (C) OR Equation of B is = log y Therefore equation of C is y = log (C) (b) Cuts -ais log = 0 = = Point is (, 0) (C) [4] 0. METHOD log 0 = log 0 log 0 y log 0 z y z log 0 y = log 0 y log 0 z = log z log 0 = log 0 log y log z y z

14 = p q r (C)(C)(C) METHOD r = 0, y = 0 p, z = 0 p log 0 0 = log 0 r y z q 0 0 r p q = log 0 r 0 = p q (A) (C)(C)(C). (a) METHOD 5 + = 5 4 A + = 4 = A N METHOD Taking logs A eg + = log 5 65, ( + )log 5 = log 65 + = log 65 log5 ( + = 4) = A N (b) METHOD Attempt to re-arrange equation (M) + 5 = a A 5 = a A N METHOD Change base to give log ( + 5) = log a (M) + 5 = a A 5 = a A N

15 . (a) e ln( + ) = e (M) + = e = e (= 8.) A N (b) log 0 (0 ) = log (accept lg and log for log 0 ) (M) = log log 500 = log A N = =.5 log 00 Note: In both parts (a) and (b), if candidates use a graphical approach, award M for a sketch, A for indicating appropriate points of intersection, and A for the answer.. (a) (i) f (a) = A N (ii) f () = 0 A N (iii) f (a 4 ) = 4 A N (b) y f f 0 Note: Award A for approimate reflection of f in y =, A for y intercept at, and A for curve asymptotic to ais. AAA N

16 4. METHOD log = log log y = log y log z = log z log + log y log z a+ b c METHOD b a c = 0, y = 0, z = 0 (C6) log b a y 0 0 log = z c b a+ c b = log0 0 = a+ c (A) 5. (a) =, (, 0), A N (b) (i) f (.999) = ln (0.00) = 6.9 A N (ii) All real numbers. A N (c) (4.64,.89) AA N 6. (a) log 5 = log 5 (M) = y (C) (b) log 5 = log 5 (M) = y (C) (c) log5 log 5 = log5 5 (M) = y (C)

17 ( ) 7. (a) f = ln A N (b) (i) Attempt to form composite (f g) () = f (ln ( + )) (M) (f g) () = e ln ( + ) = (= + ) A N (ii) Simplifying y = e In( + ) to y = + (may be seen in part (i) or later) Interchanging and y (may happen any time) eg = + y = y M (f g) () = A N 8. P log 0 = log P 0 QR QR (M) P log 0 = (log 0 P log 0 (QR )) QR (M) = (og 0 P log 0 Q log 0 R) (M) = ( y z) = y 6z or ( y z) [4] log a 5 9. (a) log 5 = (M) log a y = (C) (b) log a 0 = log a 4 + log a 5 or log a + log a 0 (M) = log a + log a 5 = + y (C) [4]

18 Part II: Solving Logarithmic Equations. Solve log + log ( ) =, for >. (Total 7 marks). (a) Let log c = p and log c 5 = q. Find an epression in terms of p and q for (i) log c 5; (ii) log c 5. (b) Find the value of d if log d 6 =.. Find the eact solution of the equation 9 = 7 ( ). Answer: Solve the equation log 7 = log 7 ( 0.4). Answer:...

19 5. Solve the equation log log log 9 = log 9. Answer:... (Total 4 marks) 6. Solve the equation 9 =. Answer:... (Total 4 marks) P 7. Let log 0 P =, log 0 Q = y and log 0 R = z. Epress log 0 in terms of, y and z. QR Answer:...(Total 4 marks)

20 8. Solve the equation 4 = Answer:... (Total 4 marks) 9. Solve the following equations. (a) log 49 = () (b) log 8 = () (c) log 5 = (d) log + log ( 7) = () (5) (Total marks) 0. (a) Given that log log ( 5) = log A, epress A in terms of. (b) Hence or otherwise, solve the equation log log ( 5) =. Answers: (a)... (b)...

21 Markscheme for Part II. recognizing log a + log b = log ab (seen anywhere) e.g. log (( )), recognizing log a b = a = b (seen anywhere) e.g. = 8 correct simplification e.g. ( ) =, 8 evidence of correct approach to solve e.g. factorizing, quadratic formula A (M) correct working A ± 6 e.g. ( 4)( + ), = 4 A N. (a) (i) log c 5 = log c + log c 5 [7] = p + q A N (ii) log c 5 = log c 5 = q A N (b) METHOD d = 6 M d = 6 A N METHOD For changing base M eg log log = d, log 0 6= log 0 d d = 6 A N. METHOD 9 =, 7 = epressing as a power of, ( ) = ( ) (M) = 4 4=

22 7 = = 7 METHOD ( ) log 9 = log 7 log 7 = = log 9 (M) (C6) 4= 7 = = 7 Notes: Candidates may use a graphical method. Award (M) for a sketch, for showing the point of intersection, for , and for. 7 (C6) 4. log 7 (( 0.4)) = l (M) 0.4 = 7 (M) = 5.4 or = 5 (G) = 5.4 (C6) Note: Award (C5) for giving both roots. 5. METHOD log log 9 + log 9 = + (M) 9 = log 9 = 9 (M) = 7 (C4) METHOD log 8 + log 9 + log 9 = log (M) = log 9 7 = 7 (C4) [4] 6. 9 = = (M) =

23 = (C4) 7. P log 0 = log P 0 QR QR (M) P log 0 = (log 0 P log 0 (QR )) QR (M) = (og 0 P log 0 Q log 0 R) (M) = ( y z) = y 6z or ( y z) 8. 4 = ( )log 0 4 = log (M) log0.565 = log 4 0 = = (C4) [4] [4] [4] 9. (a) = 49 (M) = ±7 = 7 A N (b) = 8 (M) = A N (c) = 5 (M) = 5 = A N 5 (d) log (( 7)) = (M) log ( 7) = = 8 (8 = 7) 7 8 = 0 A ( 8)( + ) = 0 ( = 8, = ) = 8 A N 0. (a) log log ( 5) = log 5 A = 5 (C) Note: If candidates have an incorrect or no answer to part (a) award (A0) []

24 (b) EITHER log = 5 = ( = ) 5 = 5 = 5 5 = OR log0 5 = log 0 if log seen in part (b). 5 (M) (M) (C4) log0 = log0 5 = 7.5 (C4)

25 Part III: Solving Eponential Functions Using Logarithms. The mass m kg of a radio-active substance at time t hours is given by m = 4e 0.t. (a) (b) Write down the initial mass. The mass is reduced to.5 kg. How long does this take? Answers: (a)... (b).... $000 is invested at 5% per annum interest, compounded monthly. Calculate the minimum number of months required for the value of the investment to eceed $000. Answer:...

26 . Michele invested 500 francs at an annual rate of interest of 5.5 percent, compounded annually. (a) (b) Find the value of Michele s investment after years. Give your answer to the nearest franc. How many complete years will it take for Michele s initial investment to double in value? () () (c) What should the interest rate be if Michele s initial investment were to double in value in 0 years? (4) (Total 0 marks) 4. A group of ten leopards is introduced into a game park. After t years the number of leopards, N, is modelled by N = 0 e 0.4t. (a) (b) How many leopards are there after years? How long will it take for the number of leopards to reach 00? Give your answers to an appropriate degree of accuracy. Give your answers to an appropriate degree of accuracy. Answers: (a)... (b)... (Total 4 marks) 5. Initially a tank contains litres of liquid. At the time t = 0 minutes a tap is opened, and liquid then flows out of the tank. The volume of liquid, V litres, which remains in the tank after t minutes is given by V = (0.9 t ). (a) (b) (c) Find the value of V after 5 minutes. Find how long, to the nearest second, it takes for half of the initial amount of liquid to flow out of the tank. The tank is regarded as effectively empty when 95% of the liquid has flowed out. Show that it takes almost three-quarters of an hour for this to happen. () () ()

27 (d) (i) Find the value of V when t = 0.00 minutes. (ii) Hence or otherwise, estimate the initial flow rate of the liquid. Give your answer in litres per minute, correct to two significant figures. () (Total 0 marks) 6. A population of bacteria is growing at the rate of.% per minute. How long will it take for the size of the population to double? Give your answer to the nearest minute. Answer:... (Total 4 marks) 7. A machine was purchased for $0000. Its value V after t years is given by V =00000e 0.t. The machine must be replaced at the end of the year in which its value drops below $500. Determine in how many years the machine will need to be replaced. Answers:...

28 8. There were 40 doctors working in a city on January 994. After n years the number of doctors, D, working in the city is given by D = n. (a) (i) How many doctors were there working in the city at the start of 004? (ii) In what year were there first more than 000 doctors working in the city? () At the beginning of 994 the city had a population of. million. After n years, the population, P, of the city is given by P = (.05) n. (b) (i) Find the population P at the beginning of 004. (ii) Calculate the percentage growth in population between January 994 and January 004. (iii) In what year will the population first become greater than million? (7) (c) (i) What was the average number of people per doctor at the beginning of 994? (ii) After how many complete years will the number of people per doctor first fall below 600? (5) (Total 5 marks)

29 Markscheme for Part III:. (a) Initial mass t = 0 mass = 4 (C) (b).5 = 4e 0.t (or 0.75 = e 0.t ) (M) ln 0.75 = 0.t (M) t = 4.90 hours (C4). 5 5% per annum = % =.5% per month (M) Total value of investment after n months, 000(.05) n > 000 (M) => (.05) n > log( ) n log (.05) > log () => n > (M) log(.05) Whole number of months required so n = 89 months. (C6) Notes: Award (C5) for the answer of 90 months obtained from using n instead of n to set up the equation. Award (C) for the answer 6 months obtained by using simple interest. Award (C) for the answer 60 months obtained by using simple interest.. (a) Value = 500(.055) (M) = = 749 (nearest franc) (b) 000 = 500(.055) t =.055 t (M) log t = =.546 log.055 It takes 4 years. (c) 000 = 500( +r) 0 or ( +r) 0 (M) = + r or log = 0 log ( + r) (M) 0 log 0 0 r = or r = 0 r = [or 7.8%] 4 [0] 4. (a) At t =, N = 0e 0.4() (M) N =. ( sf) Number of leopards =

30 (b) If N = 00, then solve 00 = 00e 0.4t 0 = e04 t ln 0 ln0 = 0.4t t = 0.4 ~ 5.76 years ( sf) [4] 5. Note: A reminder that a candidate is penalized only once in this question for not giving answers to sf (a) V(5) = 0000 (0.9 5 ) = = 7070 ( sf) (b) We want t when V = 5000 (M) 5000 = 0000 (0.9) t 0.5 = 0.9 t log (0.5) log (0.9) = t or ln (0.5) ln (0.9) = t After 0 minutes 0 seconds, to nearest second (or 600 seconds). (c) 0.05 = 0.9 t (M) log (0.05) = t = 4.97 minutes log (0.9) (M) /4 hour (AG) (d) (i) (0.9) 0.00 = 0.69 dv (ii) Initial flow rate = where t = 0, (M) dt dv 0.69 = = 69 dt 0.00 = 690 ( sf) OR dv dt = 690 (G) [0] 6..0 t = (M) ln t = ln.0 (M) = minutes (nearest minute) (C4)

31 Note: Do not accept minutes. [4] e 0.t = 500 For taking logarithms (M) 0.t ln e = ln 0.5 t = ln = 6. 7 (years) (C6) Note: Candidates may use a graphical method. Award for setting up the correct equation, (M) for a sketch, for showing the point of intersection, for 6., and for (a) (i) 40 (ii) n > 000 (M) n > (accept 6 th year or n = 6) (N) Note: Award (A0) for 000, or after 6 years, or n = 6, 000. (b) (i) (.05) 0 = 56 0 (accept or.54(million)) (ii) (M) 8.0% (accept 8.% from ) (N) (iii) (.05) n > (accept an equation) (M) n log.05 > log n > (M) 04 (accept st year or n = ) (N) 7 Note: Award (A0) for 05, after years, or n =, so 05.

32 (c) (i) = n (ii) (.05) n < 600 (M)(M) n > years (A) (N) 5 [5]