P6.5 Solving aˣ = b

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

The vast majority of learners found this question, which involved an understanding of indices and logarithms, to be accessible. Most were able to take logarithms with the same base of each side and apply the power law to both sides in one step. It was common in responses to take natural logarithms or logarithms to base 10; sometimes logarithms to base 3 or, on rarer occasions, base 7 were used. Usually, this was followed by correct division to achieve x y as the subject and gaining the accuracy mark. A few divided incorrectly and obtained the reciprocal of the correct answer and so lost the A mark. Some proceeded from a correct answer with incorrect logarithm work, such as to 7 ln 3 or ( ) ln 7 3 - , and did not gain the accuracy mark. A number of concise responses attempted to take logarithms of base 3 and achieved 3 log 7 x y = , gaining both marks. On rare occasions, responses took logarithms of base 3 and logarithms of base 7 of the original equation separately to produce 3 log 7 y x = and 7 log 3x y = and simply divided from here. Having not dealt with the powers appropriately, they left x and y variables on both sides of the equation and so scored no marks. Others took the same approach but did deal with the powers to produce 3 log 7 x y = and 7 log 3 y x = but did not recognise how to combine the equations to achieve x y as the subject by square rooting. Many learners appeared unclear about what “exact value” means and following a correct answer provided a decimal approximation. However, such responses were condoned.

Common mistakes

• take logs of the same base of both sides and use the power law: x log 3 = y log 7
• x/y = log 7 / log 3 (not the reciprocal)
• log 7/log 3 is not log(7/3) or log 7 − log 3
• a decimal alone scores 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Generally, candidates found this question accessible with many scoring consistently throughout. In part (c), candidates who had a correct equation were generally able to complete the substitution successfully. Some candidates gained full marks after having incorrectly answered part (a), by using the equation linking 10 log A and t given in the question. As it was a modelling question, units were required to score this mark; most candidates used the correct units. This part of the question was also well attempted by many candidates. Those who found difficulties often tried to use 8 t = which was information given for use in part i.

Common mistakes

• (i) 5.5 kg to the nearest 0.5 kg with units
• (ii) solve 4 = 3.162 × 1.072^t or log₁₀4 = 0.03t + 0.5: t ≈ 3.4 weeks
• using t = 8 again in (ii)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question on an exponential function saw full marks for most. Of the small number Part (b) was similarly well-answered and most used logarithms correctly to obtain a correct numerical expression for x. Those who did not use x = log4100 and arrived at log100 sometimes processed this incorrectly, arriving at log 96, log 25 or even 25. x = log 4 Some candidates who did not know how to solve the equation resorted to trial and improvement with varied success and is not a recommended strategy.

Common mistakes

• x = log₄100 = log100/log4 = 3.32
• log100/log4 is not log 25 or log100 − log4 (3.22 is a common wrong answer)
• trial and improvement not recommended

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was yet another good source of marks with the modal mark being 6 out of 6, achieved by over 35% of the candidature. Part (b) was more problematic for many candidates. There was a very good understanding of the inequality/equation to be formed, and the use of logs was almost always very good. However, a significant minority of candidates failed to gain the final A mark giving an answer of 15.3 or 16 instead of Year 17.

Common mistakes

• 20 000 × 1.08ⁿ⁻¹ > 65 000
• solve with logs: n − 1 > ln 3.25 / ln 1.08 = 15.3
• answer Year 17, not 15.3 or 16
• a sum formula is wrong

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question really highlighted the dependence on calculators amongst candidates. Part (a) required simple rationalisation of a denominator and part (b) wanted candidates to work with indices/bases. In part (a) they perhaps did not fully appreciate the wording at the top of the question ‘Solutions relying on calculator technology are not acceptable’, whilst in part (b) they did not have the algebra skills to solve on their own. In part (a), when squaring both sides many candidates squared each individual term before 2x ; they seemed happier factorising the x out when the coefficients were combining the rational numbers. In a similar vein, many instead simplified the 18 and then factored out a common 2 , leaving multiple x terms in the equation. Some candidates opted to square both sides. Where a correct quadratic was built and solved, candidates gave both roots and lost the final mark. Where x was found with an irrational denominator, most candidates then went on to rationalise the denominator, but neglected to show the steps. In many cases it was clear they had typed into the calculator and read off the simplified display. When the question says ‘not relying on calculator technology’ candidates should be showing each step to secure the marks. The 18 must also be simplified when writing in simplest form. In part (b) log base 10 or natural logs were commonly used, and these could only be evaluated on the calculator. These scored no marks. Even where attempts were made to use 1 logs to base 4 or 2 there was an over reliance on calculators to find . The most log 4 2 2 common approach was to immediately take log base 4 and simplify the left-hand side. There 3 1 - log was then often little evidence of how they got from to . 4 4 2 2 Those who chose to rewrite as common bases more often led to full marks; base 2 was most common. Whilst the use of calculator technology is encouraged on the new specification, candidates should make sure that they are confident of working without one when required. 5

Common mistakes

• write both sides as powers of 2 (2^(6x − 4) = 2^(−3/2)) or use logs base 2 or 4 with working
• log₁₀ or ln needs a calculator and scores nothing
• x = 5/12

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

As with question 1 there were many fully correct answers here. Of these, most took logs or lns of both sides, then used the power law on each side and the correct order of operations before proceeding to the correct answer of 81.6. There were a few candidates who failed to put brackets around (3p -1) but they were usually able to recover. There were several cases of candidates failing to apply the power rule to both sides, usually not dealing with the ‘210’. Some candidates misapplied the power rule, multiplying first for example by 3, rather than 3p -1. Common errors were • after applying the power rule correctly changing log5/log 4 to log(5/4). • taking either log4 of one side or log5 of the other without any application of the power law

Common mistakes

• take logs of both sides and use the power law on each side: (3p − 1)log 4 = 210 log 5
• bracket round 3p − 1
• log5/log4 is not log(5/4)
• answer with no working scores 0
• p = 81.6

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Nearly all candidates combined the equations correctly and gained the first mark. Many then struggled to progress further. A few made mistakes with the indices such as changing 3(2x) into 6x but the most common mistake was attempting to take logs incorrectly, usually log15 - log(2(2x)) or log(15 - 2x+1 ) being wrongly expanded to arrive at log15 - (x + 1)log2. Those who realised 2x+1 = 2(2x) usually went on to solve correctly, giving a correct exact answer and therefore gain full marks. Candidates need to understand that seeing an equation with an unknown index does not automatically mean that it is a question involving logarithms.

Common mistakes

• write 2^(x+1) as 2 × 2^x to get 2^x = 3
• taking logs of 15 − 2^(x+1) term by term is wrong
• exact answer log₂3

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In general, Q09 discriminated well between candidates of all abilities. Q09(b) and Q09(c) were more accessible to candidates than Q09(a). Some candidates left Q09(a) blank while others struggled to make any creditable progress. In Q09(a), many candidates wrote down a valid log equation as part of their explanation. log log log d k n V Those who used mathematical reasoning to progress from 10 10 10 n d kV to (or vice versa) usually showed an intermediate step in their working and so were successful in their explanation. A few candidates made errors in progressing between these two equations, such as an incorrect method of combining logarithms (e.g. c n log log d n kV 10 ). d V ), or an incorrect use of powers (e.g. Those who 10 10 usually did so clearly made a direct comparison of their log equation with y mx c n log k in d kV enough but sometimes did not allude to (if using 1.77 rather than 10 10 1.77 their log equation). A significant number of candidates did not apply or use k logarithms to show that 0.017. V d n d kV In Q09(b), the majority of candidates substituted and into and 30 20 n 2.08. applied logarithms correctly, leading to the correct value of A few 0.017(30)n candidates used the incorrect method of simplifying 20 to give n 20 (0.51) . Some candidates substituted in V and d the wrong way around, usually n 2.50, leading to and a few used both V and d as 20. A small minority substituted V d and into a correct log equation, while a few others correctly found n by 30 20 (log 30, log 20). finding the gradient between the known points (0, 1.77) and 10 10 2.08 (0.017) d V The most common mistake was not writing a complete equation or n log 1.77 2.08log d V 2.08. after finding a correct 10 10 V In Q09(c), most candidates applied to their exponential model and found the 60 braking distance. Some then used the incorrect method of comparing this breaking distance with the overall stopping distance. The majority attempted to calculate the thinking distance and either added it to their breaking distance d or subtracted it from 100. The main error was due to the units. Some candidates did not attempt to convert 60 kmh to ms and simply multiplied 60 by the thinking time of 0.8 seconds to 1 1 obtain a thinking distance of 48 metres. Others only managed a partial conversion, with 1 some using 1 instead of 13.3m km 75 metres as their thinking distance. In 75 some of the answers, candidates used an incorrect method of combining their values to give a dimensionally incorrect quantity. For example, they might add their breaking distance to either a speed or a time. Candidates who obtained a thinking distance of 13.3m usually reached a correct answer and conclusion. A few compared the maximum speed needed to stop at the puddle with the given 60, while even fewer n 2.08; compared the maximum value of n needed to stop at the puddle with and with correct conclusions these were valid solutions.

Common mistakes

• substitute V = 30, d = 20 the correct way round: n = 2.08
• give the complete equation d = 0.017V^2.08
• 20 = 0.017(30)ⁿ wrongly simplified to (0.51)ⁿ

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

t This question, based upon the formula modelled the value of a car in £'s against the , V ap time in years after 1st January 2001. Part (a) was very well answered by the majority of students. Most chose the correct values of t and V and successfully solved a pair of simultaneous equations. Common errors involved using incorrect values for t or failing to show that A was approximately 24 800. It is important in answering questions such as (b) that students are precise about the language and words that they choose. For example, for (b)(i) the statement was required to

Common mistakes

• t = 4 and t = 11 (years after 2001), not 5 and 12
• divide to get p⁷ = 50000/32000, p = 1.0658
• show A between 24 795 and 24 805 before ≈ 24 800

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

t This question, based upon the formula modelled the value of a car in £'s against the , V ap In part (c), students generally scored the first 3 marks by correctly using logarithms and proceeding to t =21.8 or 21.9. Most students then thought that the year must be 2023 instead of 2022 as they had rounded 21.8 to 22 and added it to 2001.

Common mistakes

• 1.0658^t = 100000/24800 so t = 21.8 or 21.9
• the year is 2022, not 2023 (rounding t to 22 and adding to 2001)