Solve equations of the form ax = b — Paper 2 · IGCSE Mathematics - Additional 0606

2 The value, V dollars, of a car aged t years is given by V = 12000 e -0.2t . (i) Write down the value of the car when it was new. [1] 2 (ii) Find the time it takes for the value to decrease to of the value when it was new. [2] 3

3 marks

This question in 0606/22 Feb/March 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

1 (a) Solve the equation 7 2x + 5 = 2.5 , giving your answer correct to 2 decimal places. [3] 5 q 3 b (b) Express ^ h 1 in the form 5a p q c, where a, b and c are constants. [3] 4 625p 12 q ^ h

6 marks

Mark scheme: Question Answer Marks Guidance 1(a)  2.5 M1 correct first anti-logging step 2 x = log 7 2.5 = 2 x + 5 or log 7  5  7  or (2 x + 5)log7 = log2.5 log 7 2.5 − 5 M1 isolates x [ x = ] 2 1  2.5 x = or log 7  5 2  7  1  log2.5  or x =  − 5  2  log7  −2.26(4...) A1 1(b) 5 B3 B1 for each term 5 2 p −3 q 4 oe If B0 then allow M1 for numerator of 3 1 125q 2 or denominator of 5p 3 q 4

This question in 0606/23 May/June 2017

2 (a) Solve 3 - 1 2 = 10 . [3] (b) Solve 2e 1 - 2y = 3e 3y + 2 . [4]

7 marks

Mark scheme: 2(a)  x  M1 Take logs: − 1 log3 = log10    2   log10  M1 Make x the subject: x = 2  + 1   log3  6.19 A1 2(b) 5 y+1 2 2 M1 for attempt to combine e = exponential terms 3 –0.281 2 M1 for taking natural logs:  2  5 y + 1 = ln    3 

This question in 0606/21 Oct/Nov 2018

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

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

8 (a) Expand ( 2 - x) 5 , simplifying each coefficient. [3] e ( 2 - )x 5 # e 80 x -x 5 (b) Hence solve 4 = e . [4] e 10x + 32

7 marks

Mark scheme: 8(a) 32 – 80x + 80x2 – 40x3 + 10x4 – x5 B3 B2 for any four or five terms correct or B1 for any three terms correct or M1 for a fully correct but unsimplified expansion 8(b) Combines powers sufficiently to be M1 able to take logs or applies correct log laws For making use of their expansion M1 from part (a) 40x2 (2 – x) [= 0] oe M1 FT their (a) if possible x = 0, x = 2 cao A1

This question in 0606/21 May/June 2020

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

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

10 The number, b, of bacteria in a sample is given by b = P + Q e t2 , where P and Q are constants and t is time in weeks. Initially there are 500 bacteria which increase to 600 after 1 week. (a) Find the value of P and of Q. [4] (b) Find the number of bacteria present after 2 weeks. [1] (c) Find the first week in which the number of bacteria is greater than 1 000 000. [3]

8 marks

Mark scheme: 10(a) P + Q = 500 and P + Q2e = 600 B1 100 2 M1 for attempt to solve by removing P Q = = 15.7 or 15.6 2 from two equations both containing 3 (e − 1) terms A1 awrt P = 484 or 485 A1 awrt 10(b) B = 484.3 + 15.65e 4 = 1338 B1 Integer value rounded down from 1338… if seen. 10(c) 2 t 1 000 000 − 484.3 M1 Make e 2 t the subject e = 15.65  1 000 000 − 484.3  M1 Take logs correctly where e 2 t > 0 or 2t = ln   n  15.65  e > 0  =t 5.5 ( 3 ) or t = 5.5 ….→ 6th week. A1 nfww

This question in 0606/22 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

5 (a) Solve the following simultaneous equations. e x + e y = 5 2 e x - 3e y = 8 [5] (b) Solve the equation e ( 2 t - 1 ) = 5e ( 5 t - 3 ) . [4]

9 marks

Mark scheme: 5(a) Solves 3e x + 3e y = 15 and 2e x − 3e y = 8 oe M1 by elimination as far as 3 e x + 2e x = 23 or substitutes e y = 5 − e x into 2e x − 3e y = 8 oe OR Solves 2e x + 2e y = 10 and 2e x − 3e y = 8 oe by elimination as far as 2e y + 3e y = 2 or substitutes e x = 5 − e y into 2e x − 3e y = 8 oe x 23 y 2 A1 e = or e = oe 5 5 x = ln4.6 [ = 1.53] oe A1 If M0 scored SC1 for using their expression of the form cex = d or y = ln0.4 [ = −0.916 ] oe d d to give x = ln provided > 0 c c Finds the other value, e y or e x , by substituting their ex M1 FT their ex or ey or ey y = ln0.4 [ = −0.916 ] oe A1 or x = ln4.6 [ = 1.53] oe 5(b) 2 t −−1 ( 5 t − 3 ) 5 t −−3 ( 2 t 1) 1 M1 e = 5 or e −= oe 5 2 − 3 t 3 2 1 A1 e = 5 or e t−= 5 1 M1 FT their e a − bt = 5 or 2 − 3t = ln5 or 3t − 2 = ln 5 ct d 1 their e −= where a, b, c and 5 d are positive integers 2 − ln5 2 + ln0.2 A1 t = or t = or 0.13[0] oe 3 3 Alternative method ln e 2 t −1 = ln5 + lne 5 t − 3 oe (M1) (2t − 1)[lne] = ln5 + (5t − 3) [ lne ] oe (A1) 5t – 2t = 3 – 1 – ln5 oe (M1) Dep on one correct log law applied with at most one sign error 2 − ln5 2 + ln0.2 (A1) t = or t = or 0.13[0] oe 3 3

This question in 0606/22 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

1 (a) Solve the equation 5 w - 1 = 12, giving your answer correct to 2 decimal places. [2] (b) Solve the equation x 2 1 3 - 5x 3 + 6 = 0. [3]

5 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) w 1 log 5 12 M1 log12 or w 1 log5 w  2.54 cao A1 1(b) Rewrites in quadratic form e.g.: M1 1 y  x 3 y 2  5 y  6  0 2 1 1 3   5 x 3  6  0 or  x and factorises or solves e.g. : M1 Factorising their 3 term (y – 2)(y – 3) = 0 quadratic x = 8 and A1 x = 27

This question in 0606/21 May/June 2022

5 (a) Solve the equation = 5 . [3] 125 x3 (b) On the axes, sketch the graph of y = 4ex + 3 showing the values of any intercepts with the coordinate axes. [2] y O x

5 marks

Mark scheme: 5(a) x 3 1 x 3 1 B1 2 4  2 1  2  25  5 2 3  5 oe or 3  25 2  25 (53 ) x  25 3 x converts the terms given to powers of 5 or 3 1 x  2  625   625 or  5 or separates the power in the numerator x 3 125 3 correctly x 1 or log625 2  log125 x 3  log5 oe or applies a correct log law 2 x 3 2 3 x 3 1 M1 FT their exponential equation in the same 5  5 oe 3 base  x  2  1 oe or their logarithmic equation with any or consistent base, 1 x 3 1 1.5 x 3 2 providing their exponential or 25  25 oe logarithmic equation has at most one sign 0.5 x 3 1 0.5 oe or arithmetic error or x 3  1  1     5 oe  5  25 3 1  x log  log125 oe 5 or x 3  1log625  x 3 log125  log5 oe 2 [ x  ] 3 3 oe A1 mark final answer; not from wrong working or 1.442249... rot to 3 or more figs. 5(b) y B2 B1 for correct shape; tending to y = 3 B1 for shape with correct curvature and correct intercept of 7 marked or (0, 7) 7 indicated y = 3 O x

This question in 0606/22 May/June 2022

1 Solve the equation 4 7x - 3 - 5 = 9 . [3]

3 marks

Mark scheme: Question Answer Marks Partial Marks 1 13 B1 x = oe 14 7 x  3 3.5 oe, soi M1 or 28 x  12 14 oe, soi 1 A1 x =  oe 14 Alternative method 196 x 2  168 x  13  0 oe (B1) factorising e.g. 14 x  13 14 x  1 (M1) 13 1 (A1) x = ,  14 14

This question in 0606/23 May/June 2022

4 Solve the following equations, giving your answers to 3 significant figures. (a) 2 3 x + 1 = 5 x - 2 [3] 2y + 1 6 (b) e = 1 + 2 y + 1 [4] e

7 marks

Mark scheme: 4(a) (3x + 1)log2 = (x – 2)log5 oe B1 (3log2 – log5)x = −log2 – 2log5 M1 FT if of equivalent difficulty x = −8.32 A1 4(b) Writes as a quadratic in e2y + 1 M1 condone one error or states u = e2y + 1 and writes as a quadratic in u oe, soi 2 y +1 ( e ) 2 − e 2 y +1 − 6  = 0 oe A1 or u 2 − u − 6 [ = 0] oe (e 2 y +1 + 2)(e 2 y +1 − 3)  = 0  leading to A1 e 2 y +1 = 3 or (u + 2)(u – 3) [=0] leading to e 2 y +1 = 3 y = 0.0493 and no other solutions A1

This question in 0606/21 Oct/Nov 2022

9 The functions f ( x) and g ( x) are defined as follows for x 2- by 3 f ( x) = x 2 + 1, g ( x) = ln ( 3 x + 2) . (a) Find fg ( x) . [1] (b) Solve the equation fg ( x) = 5 giving your answer in exact form. [3]

10 marks

Mark scheme: 9(a) (ln(3x + 2))2 + 1 oe isw B1 9(b) ln(3x + 2) = [] 2 B1 e2 = 3x + 2 M1 FT ln(3x + 2) = k, where k > 0 2e − 2 A1 x = as only solution 3 9(c) ln(3ln(3x + 2) + 2) B1 their(3ln(3x + 2) + 2) = e M1 FT their gg(x) with at most one error 3ln(3x + 2) + 2 = e A1 e − 2 M1 FT their aln(3x + 2) + b = e, where a and ln(3x + 2) = or 0.239[42…] b are non-zero constants 3 e − 2 A1 3x + 2 = e 3 or 3x + 2 = 1.270[52…] awrt −0.243 A1

This question in 0606/21 Oct/Nov 2022

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

3 (a) Solve the following simultaneous equations. 3 log 2 x + 2 log 2 y = 24 5 log 2 x - 3 log 2 y = 2 [5] 2 t + 4 (b) Solve the equation = 512 . [4] l - 2 t 2

9 marks

Mark scheme: 3(a) Correctly eliminates log 2 x or log 2 y M1 A correct equation in log 2 x only or log 2 y only x = 16 or y = 64 A2 A1 for log 2 x = 4 or log 2 y = 6 y = 64 or x = 16 A2 A1 for log 2 y = 6 or log 2 x = 4 Alternative method 3 2 24 x 5 2 (M1) x y = 2 and = 2 oe y 3 y = 64 or x = 16 (A2) A1 for y19 = 2114 oe or x19 = 276 oe x = 16 or y = 64 (A2) A1 for x3 642 = 224 oe x 5 2 or 3 = 2 oe 64 OR 163  y2 = 224 oe 16 5 2 or 3 = 2 oe y 3(b) t + 4 −−(1 2 t ) 9 B2 t + 4 −−(1 2 t ) 2 = 2 B1 for 2 = 512 or 2t + 4 = 29 +1− 2 t oe, soi 2 t + 4 9 or 1− 2 t = 2 soi 2 OR OR t + 4 − (1 − 2t ) = log 2 512 oe, soi ( t + 4 ) log a 2 − (1 − 2t )log a 2 = log a 512 log a 512 or t + 4 − (1 − 2t ) = oe log a 2 3t + 3 = 9 or better M1 t = 2 A1

This question in 0606/23 Oct/Nov 2023