2018 Paper 1 Q2

Year: 2018
Paper: 1
Question Number: 2

Course: LFM Pure and Mechanics
Section: Exponentials and Logarithms

Difficulty: 1500.0 Banger: 1488.5

Problem

If \(x=\log_bc\,\), express \(c\) in terms of \(b\) and \(x\) and prove that $ \dfrac{\log_a c}{\log_a b} = \ds \log_b c \,$.
  1. Given that \(\pi^2 < 10\,\), prove that \[ \frac{1}{\log_2 \pi}+\frac{1}{\log_5 \pi} > 2\,. \]
  2. Given that \(\ds \log_2 \frac{\pi}{\e} > \frac{1}{5}\) and that \(\e^2 < 8\), prove that \(\ln \pi > \frac{17}{15}\,\).
  3. Given that \(\e^3 >20\), \, \(\pi^2 < 10\,\) and \(\log_{10}2 >\frac{3}{10}\,\), prove that \(\ln \pi < \frac{15}{13}\,\).

Solution

\(x = \log_bc\) means that \(b^x = c\) Therefore, we can write \(\frac{\log_ac}{\log_ab} = \frac{\log_ab^{x}}{\log_ab} = \frac{x \log_ab}{\log_ab} = x = \log_bc\), giving us the change of base rule. Rearranging the chance of base rule, we get \(\frac{1}{\log_bc} = \frac{\log_ab}{\log_ac}\)
  1. Since \(\pi^2 < 10\), we have \begin{align*} \Leftrightarrow && \pi^2 &< 10 \\ \Leftrightarrow && 2\log_{10} \pi &< 1 \\ \Leftrightarrow && 2 &< \frac{1}{\log_{10} \pi} \\ \Leftrightarrow && 2 &< \frac{\log_{10} 2 + \log_{10} 5}{\log_{10} \pi} \\ \Leftrightarrow && 2 &< \frac{\log_{10} 2}{\log_{10} \pi} + \frac{\log_{10} 5}{\log_{10} \pi} \\ \Leftrightarrow && 2 &< \frac1{\log_2\pi}+ \frac1{\log_5\pi} \\ \end{align*}
  2. Since \(e^2 < 8 \Rightarrow 2 < 3 \ln 2 \Rightarrow \ln 2 > \frac23\) \begin{align*} && \log_2 \frac{\pi}{e} &= \frac{\ln \frac{\pi}{e}}{\ln 2} \\ &&&= \frac{\ln \frac{\pi}{e}}{\ln 2} \\ &&&< \frac{3(\ln \pi - 1)}{2} \\ \Rightarrow && \frac{1}{5} &< \frac{3 \ln \pi - 1}{2} \\ \Rightarrow && \frac{2}{15} + 1 = \frac{17}{15}&< \ln \pi \end{align*}
  3. From the inequalities given we can set up two linear inequalities in \(\ln 2\) and \(\ln 5\) \begin{align*} && 20 &< e^3 \\ \Rightarrow && 2\ln 2 + \ln 5 &< 3 \tag{*}\\ \\ && \frac{3}{10} &< \log_{10} 2 \\ \Rightarrow && \frac{3}{10} &< \frac{\ln 2}{\ln 10} \\ \Rightarrow && 3 \ln 2 + 3 \ln 5 &< 10 \ln 2 \\ \Rightarrow && -7 \ln 2 + 3 \ln 5 &< 0 \tag{**}\\ \\ \\ && \pi^2 & < 10 \\ \Rightarrow && 2 \ln \pi &< \ln 2 + \ln 5 \tag{***}\\ \end{align*} It would be nice to combine \((*)\) and \((**)\) to bound \(\ln 2+ \ln 5\), so we want to use a linear combination such that \begin{align*} && \begin{cases} 2x -7y &= 1 \\ x + 3y &= 1\end{cases} \\ \Rightarrow && \begin{cases} y &= \frac1{13} \\ x &= \frac{10}{13}\end{cases} \\ \\ \Rightarrow && \ln 2 + \ln 5 < \frac{30}{13} + 0 \\ \Rightarrow && \ln \pi < \frac{15}{13} \end{align*}
Examiner's report
— 2018 STEP 1, Question 2
Mean: ~10.3 / 20 (inferred) Second Most Popular Inferred 10.3/20 from 'just over half marks' (half=10, 'just over' → +0.3).

This question highlighted that most people had less than an adequate understanding of inequalities. Many students decided to turn the given statements into equalities (such as 10 and work through before finally, optimistically, inserting an inequality at the end. Others reversed inequalities when taking logs. Only a tiny number considered why taking logs of an inequality was acceptable. Although there were many possible choices for the base of logarithms to work with, blindly manipulating things until the result appears was unlikely to get to the correct answer. Too many candidates produced a "stream of consciousness" including many correct statements but without any sense of creating an argument. Mathematics should be about communicating ideas. In particular, it is clear that many candidates did not know that it is flawed to start at the required result and manipulate until a known theorem is reached. Particularly when working with inequalities, these types of arguments are not always reversible. Nonetheless, Q2 was the second most popular question and with the second highest mean score (just over half marks); this was mostly as a result of being given so much information upfront.

In order to get the fullest picture, this document should be read in conjunction with the question paper, the marking scheme and (for comments on the underlying purpose and motivation for finding the right solution-approaches to questions) the Hints and Solutions document. The purpose of the STEPs is to learn what students are able to achieve mathematically when applying the knowledge, skills and techniques that they have learned within their standard A-level (or equivalent) courses … but seldom within the usual range of familiar settings. STEP questions require candidates to work at an extended piece of mathematics, often with the minimum of specific guidance, and to make the necessary connections. This requires a very different mind-set to that which is sufficient for success at A-level, and the requisite skills tend only to develop with prolonged and determined practice at such longer questions. One of the most crucial features of the STEPs is that the routine technical and manipulative skills are almost taken for granted; it is necessary for candidates to produce them with both speed and accuracy so that the maximum amount of time can be spent in thinking their way through the problem and the various hurdles and obstacles that have been set before them. Most STEP questions begin by asking the solver to do something relatively routine or familiar before letting them loose on the real problem. Almost always, such an opening has not been put there to allow one to pick up a few easy marks, but rather to point the solver in the right direction for what follows. Very often, the opening result or technique will need to be used, adapted or extended in the later parts of the question, with the demands increasing the further on that one goes. So a candidate should never think that they are simply required to 'go through the motions'; rather they will, sooner or later, be required to show either genuine skill or real insight in order to make a reasonably complete effort. The more successful candidates are the ones who manage to figure out how to move on from the given starting-point. Finally, reading through a finished solution is often misleading – even unhelpful – unless you have attempted the problem for yourself. This is because the thinking has been done for you. So, when you read through the report and look at the solutions (either in the mark scheme or the Hints and Solutions booklet), try to figure out how you could have arrived at the solution, learn from your mistakes and pick up as many tips as you can whilst working through past paper questions. This year far too many candidates wasted time by attempting more than six questions, with many of these candidates picking up 0-4 marks on several 'false starts' which petered out the moment some understanding was required. There were almost 2000 candidates for this SI paper. Almost one-sixth of candidates failed to reach a total of 30 and around two-thirds fell below half-marks overall. This highlights the fact that many candidates don't find this test an easy one. At the other end of the spectrum, almost one-in-ten managed a total of 84 out of 120 – these candidates usually marked out by their ability to complete whole questions – with almost 4% of the entry achieving the highly praiseworthy feat of getting into three-figures with their overall score. The paper is constructed so that question 1 is very approachable indeed, the intention being to get everyone started with some measure of success; unsurprisingly, Q1 was the most popular question of all, with almost all candidates attempting it, and it also turned out to be the most successful question on the paper with a mean score of more than 15 out of 20. Around 7% of candidates didn't make any kind of attempt at it at all. In order of popularity, Q1 was followed by Qs. 2, 7, 4 and 3. Indeed, it was the pure maths questions in Section A that attracted the majority of attention from candidates, with the most popular applied question (Q9, mechanics) still getting fewer 'hits' than the least popular pure question (Q5). Questions 10, 11 and 13 proved to attract very little attention from candidates and many of the attempts were minimal.

Source: Cambridge STEP 2018 Examiner's Report · 2018-p1.pdf
Rating Information

Difficulty Rating: 1500.0

Difficulty Comparisons: 0

Banger Rating: 1488.5

Banger Comparisons: 1

Show LaTeX source
Problem source
If $x=\log_bc\,$, express $c$ in terms of $b$ and $x$
and prove that
$
\dfrac{\log_a c}{\log_a b}
=
\ds \log_b c 
\,$.
\begin{questionparts}
\item Given that $\pi^2 < 10\,$, prove that
\[
\frac{1}{\log_2 \pi}+\frac{1}{\log_5 \pi} > 2\,.
\]
\item Given that $\ds \log_2  \frac{\pi}{\e}  > \frac{1}{5}$ 
and that $\e^2 < 8$, prove that $\ln \pi > \frac{17}{15}\,$.
\item Given that $\e^3 >20$, \,
$\pi^2 < 10\,$ 
 and  $\log_{10}2 >\frac{3}{10}\,$, 
prove that $\ln \pi < \frac{15}{13}\,$.
\end{questionparts}
Solution source
$x = \log_bc$ means that $b^x = c$

Therefore, we can write $\frac{\log_ac}{\log_ab} = \frac{\log_ab^{x}}{\log_ab} = \frac{x \log_ab}{\log_ab} = x = \log_bc$, giving us the change of base rule.

Rearranging the chance of base rule, we get $\frac{1}{\log_bc} = \frac{\log_ab}{\log_ac}$

\begin{questionparts}
\item Since $\pi^2 < 10$, we have

\begin{align*}
\Leftrightarrow && \pi^2 &<  10 \\
\Leftrightarrow && 2\log_{10} \pi &<  1 \\
\Leftrightarrow && 2 &<  \frac{1}{\log_{10} \pi}  \\
\Leftrightarrow && 2 &<  \frac{\log_{10} 2 + \log_{10} 5}{\log_{10} \pi}  \\
\Leftrightarrow && 2 &<  \frac{\log_{10} 2}{\log_{10} \pi} + \frac{\log_{10} 5}{\log_{10} \pi}  \\
\Leftrightarrow && 2 &< \frac1{\log_2\pi}+ \frac1{\log_5\pi} \\
\end{align*}

\item Since $e^2 < 8 \Rightarrow 2 < 3 \ln 2 \Rightarrow \ln 2 > \frac23$
\begin{align*}
&& \log_2 \frac{\pi}{e} &= \frac{\ln \frac{\pi}{e}}{\ln 2} \\
&&&= \frac{\ln \frac{\pi}{e}}{\ln 2} \\
&&&< \frac{3(\ln \pi - 1)}{2} \\
\Rightarrow && \frac{1}{5} &< \frac{3 \ln \pi - 1}{2} \\
\Rightarrow && \frac{2}{15} + 1 = \frac{17}{15}&< \ln \pi
\end{align*}


\item From the inequalities given we can set up two linear inequalities in $\ln 2$ and $\ln 5$
\begin{align*}
&& 20 &< e^3 \\
\Rightarrow && 2\ln 2 + \ln 5 &< 3 \tag{*}\\
\\
&& \frac{3}{10} &< \log_{10} 2 \\
\Rightarrow && \frac{3}{10} &< \frac{\ln 2}{\ln 10} \\
\Rightarrow && 3 \ln 2 + 3 \ln 5 &< 10 \ln 2 \\
\Rightarrow && -7 \ln 2 + 3 \ln 5 &< 0 \tag{**}\\
\\
\\
&& \pi^2 & < 10 \\
\Rightarrow && 2 \ln \pi &< \ln 2 + \ln 5 \tag{***}\\
\end{align*}

It would be nice to combine $(*)$ and $(**)$ to bound $\ln 2+ \ln 5$, so we want to use a linear combination such that

\begin{align*}
&& \begin{cases} 2x -7y &= 1 \\ 
x + 3y &= 1\end{cases}  \\
\Rightarrow && \begin{cases} y &= \frac1{13} \\ 
x  &= \frac{10}{13}\end{cases} \\
\\
\Rightarrow && \ln 2 + \ln 5 < \frac{30}{13} + 0 \\
\Rightarrow && \ln \pi < \frac{15}{13}
\end{align*}

\end{questionparts}