| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2017 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Two unrelated log parts: both solve equations |
| Difficulty | Moderate -0.3 This is a straightforward application of logarithm laws (subtraction rule and power rule) with standard algebraic manipulation. Part (i) requires combining logs and solving a quadratic, while part (ii) involves rearranging to express one variable in terms of another. Both are routine textbook exercises requiring no novel insight, making this slightly easier than average for A-level. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2\log_{10}(x-2)-\log_{10}(x+5)=0 \Rightarrow \log_{10}(x-2)^2=\log_{10}(x+5)\) | M1 | Use of power law of logs |
| For 'undoing' logs: setting \(\log_{10}...=\log_{10}...\) or using subtraction law and \(0=\log_{10}1\) | M1 | |
| \((x-2)^2=(x+5) \Rightarrow x^2-5x-1=0\) | A1 | Correct simplified quadratic \(x^2-5x-1=0\) |
| \(x=\frac{5\pm\sqrt{29}}{2} \Rightarrow x=\frac{5+\sqrt{29}}{2}\) only | M1, A1 | M1: correct attempt to solve 3TQ of equivalent difficulty (formula, completing square, calculator); A1: cso \(\frac{5+\sqrt{29}}{2}\) or exact simplified equivalent, without extra answers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\log_p(4y+1)-\log_p(2y-2)=1 \Rightarrow \log_p\!\left(\frac{4y+1}{2y-2}\right)=\log_p p\) | M1, M1 | First M1: use of subtraction (or addition) law of logs; second M1: using \(1=\log_p p\) or equivalent to get equation not involving logs; note: \((4y+1)-(2y-2)=p\) implies M0M1 |
| \(\left(\frac{4y+1}{2y-2}\right)=p\) | A1 | Correct equation in \(p\) and \(y\) not involving logs; accept \(\left(\frac{4y+1}{2y-2}\right)=p^1\) |
| \(4y+1=2py-2p \Rightarrow y=\frac{1+2p}{2p-4}\) | M1, A1 | M1: attempt to change subject (must include cross multiplication, collection of terms in \(y\), factorisation of \(y\) term); A1: cso \(y=\frac{1+2p}{2p-4}\) or equivalent e.g. \(y=\frac{-1-2p}{4-2p}\) |
# Question 9 (Logarithms):
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2\log_{10}(x-2)-\log_{10}(x+5)=0 \Rightarrow \log_{10}(x-2)^2=\log_{10}(x+5)$ | M1 | Use of power law of logs |
| For 'undoing' logs: setting $\log_{10}...=\log_{10}...$ or using subtraction law and $0=\log_{10}1$ | M1 | |
| $(x-2)^2=(x+5) \Rightarrow x^2-5x-1=0$ | A1 | Correct simplified quadratic $x^2-5x-1=0$ |
| $x=\frac{5\pm\sqrt{29}}{2} \Rightarrow x=\frac{5+\sqrt{29}}{2}$ only | M1, A1 | M1: correct attempt to solve 3TQ of equivalent difficulty (formula, completing square, calculator); A1: cso $\frac{5+\sqrt{29}}{2}$ or exact simplified equivalent, without extra answers | **(5)** |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\log_p(4y+1)-\log_p(2y-2)=1 \Rightarrow \log_p\!\left(\frac{4y+1}{2y-2}\right)=\log_p p$ | M1, M1 | First M1: use of subtraction (or addition) law of logs; second M1: using $1=\log_p p$ or equivalent to get equation not involving logs; note: $(4y+1)-(2y-2)=p$ implies M0M1 |
| $\left(\frac{4y+1}{2y-2}\right)=p$ | A1 | Correct equation in $p$ and $y$ not involving logs; accept $\left(\frac{4y+1}{2y-2}\right)=p^1$ |
| $4y+1=2py-2p \Rightarrow y=\frac{1+2p}{2p-4}$ | M1, A1 | M1: attempt to change subject (must include cross multiplication, collection of terms in $y$, factorisation of $y$ term); A1: cso $y=\frac{1+2p}{2p-4}$ or equivalent e.g. $y=\frac{-1-2p}{4-2p}$ | **(5)(10 marks)** |9. (i) Find the exact value of $x$ for which
$$2 \log _ { 10 } ( x - 2 ) - \log _ { 10 } ( x + 5 ) = 0$$
(ii) Given
$$\log _ { p } ( 4 y + 1 ) - \log _ { p } ( 2 y - 2 ) = 1 \quad p > 2 , y > 1$$
express $y$ in terms of $p$.\\
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\hfill \mbox{\textit{Edexcel C12 2017 Q9 [10]}}