Know and use the laws of logarithms — Paper 2 · IGCSE Mathematics - Additional 0606

6 (a) (i) Express - 8x 9 x -3 in the form axb, where a and b are constants to be found. [2] 3 9 6 -3 (ii) Hence solve the equation - 8x x =- 6250 . [2] ` `j j (b) It is given that y = log a ( ax) + 2 log a ( 4x - 3) - 1, where a is a positive integer. (i) Explain why x must be greater than 0.75. [1] - 24x + 9x) . [3] (ii) Show that y can be written as log a ( 16x 3 2 (iii) Find the value of x for which y = log a ( 9x) . [2]

10 marks

This question in 0606/22 Feb/March 2017

5 (i) Show that [ 0. 4x 5 ( 0. 2 - ln 5 x)] = kx 4 ln 5x , where k is an integer to be found. [2] d x (ii) Express ln 125x 3 in terms of ln 5x . [1] ( x 4 ln 125 x 3) d x . [2] (iii) Hence find y

5 marks

Mark scheme: their 2 x 5(i) ( 4 5  − 5  M1 clearly applies correct form of product ) (0.2 − ln5 x ) + 0.4 x  their  oe or rule  5 x  4  4 5  5   oe their 0.4 x −  ( their 2 x ) ln 5 x + 0 .4 x  their     5 x   − 2 x 4 ln5 x isw A1 nfww 5(ii) 3ln5x or ln5 x + ln5 x + ln5 x B1 −2 x ln5 x ) dx oe 3 45(iii) −∫32 ( 4 M1 FT k = 2 from (i) allow for ( 2 x ln5 x ) dx 2 ∫ or, when k = −2, for 5 (0.2 − ln5 x ) x 4 ln5 x ) dx = −0.2 x ∫ ( 2 4 5 or − 3 x ln5 x ) dx = 0.4 x (0.2 − ln5 x ) oe 3 ∫ ( or, when FT k = 2, for 5 (0.2 − ln5 x ) x 4 ln5 x ) dx = 0.2 x ∫ ( 2 4 5 or 3 x ln5 x ) dx = 0.4 x (0.2 − ln5 x ) oe 3 ∫ ( 3 5 A1 nfww; implies M1 − ( 0.4 x (0.2 − ln5 x ) )[ + c ] oe isw cao 5 2 An answer of 0.6 x (0.2 − ln5 x ) following k = 2 from (i) implies M1 A0

This question in 0606/22 May/June 2017

7 (a) Given that a 7 = b , where a and b are positive constants, find, (i) log a b , [1] (ii) log b a . [1] 1 (b) Solve the equation log 81 y =- . [2] 4 32 x 2 - 1 (c) Solve the equation 2 = 16 . [3] 4 x

7 marks

Mark scheme: 7(a)(i) 7 B1 7(a)(ii) 1 1 B1 FT their 7 must not be 1 if following or through 7 their 7 7(b) − 1 − 1 M1 Anti-logs y = 81 4 or y = 3−1 or y = 9 2 oe 1 A1 nfww; implies the M1; y = only or 0.333[3….] only y = …. must be seen at least once 3 1 4 1 If M0 then SC1 for e.g. 81−= as final 3 answer 7(c) 2 5( x 2 −1) 2 B1 converts the terms given left hand side to 2 5( x −1) 4 2 32 x × 32 −1 powers of 2 or 4; may have cross- 2 oe or 2 oe or 2 (2 2 ) x 4 x 4 x multiplied or log32 x 2 −−1 log4 x 2 = log16 oe or separates the power in the numerator correctly or applies a correct log law 3 x 2 5 2 M1 combines powers and takes logs or 2 −= 16 oe ⇒ 3 x − 5 = 4 oe 3 2 5 equates powers; x 2 2 3 2 5 or 4 −= 16 oe ⇒ x − = 2 oe 2 2 or brings down all powers for an equation 8 x 2 2 already in logs or = 16 oe ⇒ x log8 = log512 oe 32 2 2 condone omission of necessary brackets or ( x − 1)log32 − x log 4 = log16 oe for M1; condone one slip [ x = ] ± 3 isw cao A1 or ± 1.732050... rot to 3 or more figs. isw

This question in 0606/22 May/June 2017

4 Solve the simultaneous equations log 2 x + 4 = 2 log 2 y , ^ h log 2 7y - x = 4 . ^ h [5]

5 marks

Mark scheme: 4 x + 4 = y 2 B1 7 y − x = 16 B1 allow 2 4 for1 6 7 y − 16 + 4 = y 2 y 2 − 7 y + 12 → ( y − 3 )( y − 4 )( = 0 ) M1 Attempt to eliminate x or y to 2 obtain a three term quadratic. or x − 17 x + 60 → ( x − 5 )( x − 12 )( = 0 ) Solve a three term quadratic M1 M1dep → y = 3, x = 5 or y = 4 x = 12 A1 Allow for values seen even if correct pairs not clear.

This question in 0606/22 Oct/Nov 2017

4 Solve the simultaneous equations log 3 x + 1 = 1 + log 3 y , ^ h log 3 x - y = 2 . [5] ^ h

5 marks

Mark scheme: 4 log 3 3 = 1 or log 3 9 = 2 B1 implied by one correct equation x + 1 = 3 y B1 x − y = 9 B1 solve correct equations for x or y M1 x = 14 and y = 5 A1

This question in 0606/23 Oct/Nov 2017

4 Solve (i) 2 3x - 1 = 6 , [3] 2 (ii) log 3 (y + 14) = 1 + . [5] log y 3

8 marks

Mark scheme: 4(i) Take logs : ( 3 x − 1) log2 = log6 M1 log6 A1 + 1 log2 Make x the subject : x = oe 3 awrt 1.19 A1 or awrt 1.195 4(ii) 1 = log 3 3 B1 2 B1 = 2log 3 y log y 3 3 y 2 − y − 14 = 0 B1 ( 3 y − 7 )( y + 2 ) = 0 M1 Solve a three term quadratic 7 A1 y = only 3

This question in 0606/22 Oct/Nov 2018

6 Solve the simultaneous equations log 2 (x + 2y) = 3 , log 2 3x - log 2 y = 1. [5]

5 marks

Mark scheme: 6 3 x B1 implied by one correct equation log28 = 3 or log3 x − log y = log (any base) y or log22 = 1 soi x + 2 y = 8 B1 3 x B1 = 2 y solve correct equations for x or y M1 x = 2 and y = 3 A1

This question in 0606/23 Oct/Nov 2018

3 (a) Solve e 2x + 1 = 3e 4 - 3x . [3] (b) Solve lg (y - 6) + lg (y + 15) = 2 . [5]

8 marks

Mark scheme: 3(a) obtain e5x – 3 = 3 M1 OR Take logs → 2x + 1 = ln3 + 4 – 3x take logs correctly M1 OR Collect like terms → 5x = 3 + ln3 → 5x – 3 = ln3 3 + ln3 A1 x = or x = 0.820 5 3(b)` Use of laws of logs M1 → lg(y – 6)(y + 15) = 2 Uses 10ଶ= 100 B1 → [(y – 6)(y +15)] = 100 Obtain correct quadratic A1 → y2 + 9y – 190 = 0 Solve a three term quadratic M1 y = 10 only A1

This question in 0606/21 Oct/Nov 2019

9 (a) Solve the equation x - 2 = 243. [3] 27 1 (b) log a b - = log b a, where a 2 0 and b 2 0. 2 Solve this equation for b, giving your answers in terms of a. [5]

8 marks

Mark scheme: 9(a) 310 x B1 3 x− 6 [ = 243] oe or 3 log 9 5 x − log 27 x− 2 = log 243 oe 37 x+ 6 = 35 soi oe or M1 5 x ( log9 ) − ( x − 2 ) log27 = log243 1 A1 x = − 7 9(b) 1 1 1 B2 1 log a b − = B1 for bringing down the power of 2 2 log a b 2 1 1 e.g. log a b or for a change of base 1 2 2 or − = log b a log b a 2 1 e.g. loga b Clears the fraction and rearranges M1 1 2 1 ( log a b ) − log a b = 1 oe 2 2 ( log a b ) 2 − log a b − 2 = 0 oe or let x = log a b x 2 − x − 2 = 0 oe or 1 1 2 − log b a = (log b a ) 2 2 0 = 2(log b a ) 2 + log b a − 1 oe or let y = log b a 2 y 2 + y −=1 0 (log a b − 2)(log a b + 1) oe or M1 (2logb a − 1)(logb a + 1) [log a b = 2, log a b = −1 or A1 1 log b a = , log b a = −1 2 leading to ] b = a2, b = oe

This question in 0606/22 May/June 2020

3 Write 3 lg x + 2 - lg y as a single logarithm. [3]

3 marks

Mark scheme: 3 Uses lg100 = 2 or 3lgx = lgx 3 . B1 a  B1 Uses lg a + lg b = lg ab or lg a − lg b = lg   b  lg 100 x 3  B1 Correct final answer  y 

This question in 0606/21 Oct/Nov 2020

8 The population P, in millions, of a country is given by P = A # bt , where t is the number of years after January 2000 and A and b are constants. In January 2010 the population was 40 million and had increased to 45 million by January 2013. (a) Show that b = 1.04 to 2 decimal places and find A to the nearest integer. [4] (b) Find the population in January 2020, giving your answer to the nearest million. [1] (c) In January of which year will the population be over 100 million for the first time? [3]

8 marks

Mark scheme: 8(a) 40 = A × b10 and 45 = A × b13 B1 3 45 M1 Divide to find b 3. b = 40 b = 1.04 A1 A = 27 A1 8(b) 59 B1 P = 27 × 1.04 20 8(c) 100 = 27 × 1.04 t M1 Insert P = 100 in their expression  100  M1 Rearrange to make t the subject log    27  t = oe log1.04 t = 33.4 → Year 2034 A1

This question in 0606/21 Oct/Nov 2020

4 Solve the simultaneous equations. log 3 ( x + y) = 2 2 log 3 ( x + 1) = log 3 ( y + 2) [6]

6 marks

Mark scheme: 4 x + y = 9 B1 (x + 1)2 = y + 2 B1 x + (x + 1)2 – 2 = 9 M1 Replace y or x. Allow unsimplified using their three term expressions both or (10 – y)2 = y + 2 containing x and y terms. Condone one sign or arithmetic error. Result must be a quadratic function. x2 + 3x – 10 (= 0) A1 Correct 3 term quadratic or y2 – 21y + 98 (= 0) x = –5 and x = 2 M1 Dep on correct method to solve their or y = 7 and y = 14 quadratic or (x + 5)(x – 2) or (y – 7)(y – 14) x = 2 and y = 7 only A1 Reject x = –5, y = 14 as log –4 is not appropriate

This question in 0606/22 Oct/Nov 2020

8 DO NOT USE A CALCULATOR IN THIS QUESTION. log 2 ( y + 1) = 3 - 2 log 2 x log 2 ( x + 2) = 2 + log 2 y (a) Show that x 3 + 6x 2 - 32 = 0 . [4] (b) Find the roots of x 3 + 6x 2 - 32 = 0 . [4] (c) Give a reason why only one root is a valid solution of the logarithmic equations. Find the value of y corresponding to this root. [2]

10 marks

Mark scheme: 8(a) x 2 ( y + 1) = 8 oe B1 x + 2 = 4 y oe B1 2  x + 2  M1 eliminate y from correct equations x  + 1  = 8  4  x 3 + 6 x 2 − 32 = 0 A1 answer given 8(b) x = 2 or x = −4 seen B1 or ( x − 2 ) or ( x + 4 ) seen find quadratic factor M1 x2 and 16 or long division to x2 + kx or x2 and –8 or long division to x2 + kx not from expanding two linear factors 2 2 A1 x + 8 x + 16 or x + 2 x − 8 ( ) ( ) 2 A1 answer only without working earns B1 ( x − 2)( x + 4) and x = 2, − 4, − 4 above only 8(c) no real value for log 2 ( −4 ) B1 must identify specific term in one of original equations and use x = –4 or log 2 ( −+4 2 ) y = 1 B1

This question in 0606/23 Oct/Nov 2020

7 (a) Use logarithms to solve the following equation, giving your answer correct to 1 decimal place. 5 x - 2 = 3 # 2 2 x + 3 [4] (b) Solve the equation log 3 (y 2 + 11 ) - 2 = log 3 ( y - 1 ) . [5]

9 marks

Mark scheme: 7(a) x−2 2 x+3 M1 log5 = log3 + log2 soi ( x − 2)log5 = log3 + (2 x + 3)log 2 oe M1 dep on previous M1; Condone one sign or bracketing error log 3 + 3log 2 + 2 log 5 A1 x = soi log 5 − 2 log 2 x = 28.7 A1 7(b)  y 2 + 11  B1 log 3   = log 3 ( y − 1)  9   y 2 + 11  or log 3   = 2 oe  y − 1  y 2 + 11 y 2 + 11 M1 = y − 1 or = 9 oe 9 y − 1 2 A1 y − 9y + 20 = 0 Solves their 3-term quadratic M1 dep on previous M1 y = 4, y = 5 A1

This question in 0606/21 Oct/Nov 2021

4 (a) Solve the equation log 6 ( 2x - 3) = . Give your answer in exact form. [2] 2 (b) Solve the equation ln 2u - ln ( u - 4) = 1. Give your answer in exact form. [3] 3 v (c) Solve the equation 2 v - 5 = 9 . [3] 27

8 marks

Mark scheme: 4(a) 1 M1 2 x − 3 = 6 2 oe, soi 1 A1 6 2 + 3 6 + 3 x = or x = 2 2 4(b) 2u 2u M1 Condone one sign or bracketing ln = lne soi or ln = 1 soi error u − 4 u − 4 or ln 2u = lne(u − 4) soi 2u M1 FT their logarithmic equation = e or 2u = e (u – 4) oe u − 4 4e −4e A1 u = or u = or equivalent exact form e − 2 2 − e 4(c) v B1 3v 2 9 2 2 v −=5 3 oe soi or 2 v −=5 9 oe soi ( 33 )  32   9      or log3v − log27 2 v − 5 = log9 oe soi 15 − 5v = 2 oe or v log3 − (2 v − 5)log27 = log9 M1 FT their exponential equation in the same base or their logarithmic equation with any consistent base, providing their exponential or logarithmic equation has at most one sign or arithmetic error 13 A1 v = oe 5

This question in 0606/23 Oct/Nov 2021

2 (a) Write 2 lg x - lg ( x + 6) + lg 3 as a single logarithm to base 10. [2] (b) Hence solve the equation 2 lg x - lg ( x + 6) + lg 3 = 0 . [4] ` j

6 marks

Mark scheme: 2(a) x 2 B2 B1 for any two log laws lg oe, nfww applied correctly e.g. 3( x  6) x 2 lg  lg3 x  6 2(b) x 2 B1 x 2 lg  lg1 FT their lg 3( x  6) 3( x  6) 0 x 2 providing a single logarithm or 10  3( x  6) x 2  3 x  18  0 B1 dep on B2 in part (a) Factorises or solves their 3-term quadratic M1 x = 6 indicated as only solution A1 dep on B2 in part (a)

This question in 0606/21 May/June 2022

7 (a) (i) Write down the set of values of x for which lg ( 5x - 3) exists. [1] (ii) Solve the equation lg ( 5x - 3) = 1. [1] 1 (b) It is given that log y x = 4 + log y 64 + log y 162 , where y 2 0 . Find an expression for y in 2 terms of x. Simplify your answer. [5]

7 marks

Mark scheme: 7(a)(i) x  0.6 B1 7(a)(ii) B1 nfww  x  13 or 2.6 5 7(b) 1 B1 log y 64  log y 8 soi 2 or 4 = log y y 4 soi x B2 B1 for one further correct application Combines log terms e.g. log y  4 of a relevant log law 1296 e.g. log y x  4  log y 1296 4 1296 or log y x  log y 1296 y or log y  4 4 x or log y x  log y y  log y 8  log y 162 4 x 4 M1 y  or 1296y  x oe 1296 1 A1 mark final answer 4 x x 4 y  or y  6 6

This question in 0606/23 May/June 2022

5 Variables P and T are known to be connected by the relationship P = AbT , where A and b are constants. Values of P are found for certain values of time, T. (a) Show that a graph of lgP against T will be a straight line. [2] (b) lg P 12 10 8 6 4 2 T 0 2 4 6 8 10 12 14 The diagram shows the graph of lgP against T. The graph passes through (0, 6) and (14, 12). Find the values of A and b. [4] (c) Using the graph or otherwise, find the length of time for which P is between 100 million and 1000 million. [3]

9 marks

Mark scheme: 5(a) lg P  lg A  T lg b oe nfww B2 Must be seen and not from wrong working and correct comparison with y = mx + c soi B1 for lg P  lg A  T lg b isw, nfww 5(b) A = 106 oe isw and 4 B2 for A = 106 oe isw 3 or B1correct method which could be used b = 10 7 oe isw to find A e.g. 3 lgA = 6 or 12 =  14  lg A 7 3 B2 for b = 10 7 oe isw or B1correct method which could be used to find b e.g. 12  6 lgb = oe or 12 = 14lg b  6 14  0 5(c) lg P1 = 8 and lg P2 = 9 soi M1 If graph not used then allow M1 for substitution of their A and their b in the leading to exponential equation as far as 108 10 9  (theirb )T and  (theirb )T T1 = 4.6 to 4.8 theirA theirA or T2 = 6.8 to 7.2 OR substitution of their A and their b or their lgA and their lg b in the log equation lg108  their lg A  T (their lg b ) or better and lg109  their lg A  T (their lg b ) or better Difference of correct times: M1 T2  T1 where T2 = 6.8 to 7.2 T1 = 4.6 to 4.8 Answer in range 2.2 to 2.4 nfww A1 Alternative method 9  8 (M2) M1 for lg108 = 8 and lg109 = 9 and Change in T = 3 Change in lg P 3  7 Change in T 7 Answer in range 2.2 to 2.4 nfww (A1)

This question in 0606/22 May/June 2023

4 Solve the following equations. x + 1 2 ( e ) (a) = 10 [4] x e 1 (b) 2 log 9 y - log 9 ( 4y - 9) = [5] 2

9 marks

Mark scheme: 4(a) e 2 x + 2 B1 = 10 oe, soi x e 2 e1.5 x + 2 = 10 oe M1 e 2 x + k e kx + 2 FT = 10 oe or = 10 oe x x e 2 e 2 where k is an integer and k > 0 e 2 x + 2 or = 10 oe x e n where n is an integer and n > 1 or n = –2 1.5 x + 2 = ln10 oe M1 FT an expression of, or equivalent to, the forme ax + b = 10 oe where a and b are non-zero constants 2 A1 x = ( ln10 − 2 ) oe, isw or 0.202 3 or 0.2017[23…] rot to 4 or more dp isw 4(b) 2 1 M2 M1 for at least one correct log law used in a y 2 = 9 nfww correct equation e.g. 4 y − 9 y 2 1 2 2 1 or log 9 = log 9 9 oe log 9 y − log 9 (4 y − 9) = 4 y − 9 2 y 2 1 or log 9 = 4 y − 9 2 1 or 2log 9 y − log 9 (4 y − 9) = log 9 9 2 y 2 − 12 y + 27[ = 0] nfww A1 ( y − 3 )( y − 9 ) = 0 DM1 dep on at least M1 previously awarded y = 3, y = 9 nfww A1

This question in 0606/22 Oct/Nov 2023