| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2024 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Solve by showing reduces to polynomial |
| Difficulty | Standard +0.3 This is a standard A-level logarithm question requiring systematic application of log laws (power rule, addition rule) and change of base, followed by routine polynomial factorization. Part (a) is guided ('show that'), part (b)(i) uses the factor theorem with a given root, and part (b)(ii) requires checking domain restrictions. While multi-step, it follows a predictable template with no novel insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.06f Laws of logarithms: addition, subtraction, power rules |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2\log_4(x+3) = \log_4(x+3)^2\) or \(\frac{1}{2} = \log_4 2\) | B1 | May be implied by further work |
| Combines two terms e.g. \(2\log_4(x+3) + \log_4 x = \log_4 x(x+3)^2\) | M1 | Must correctly combine at least two original terms; \(2\log_4(x+3) + \log_4 x = 2\log_4 x(x+3)\) is M0 |
| e.g. \(x(x+3)^2 = 2(4x+2)\) | A1 | Correct intermediate equation not involving logs (but not the given answer) |
| e.g. \(x(x^2+6x+9) = 8x+4 \Rightarrow x^3+6x^2+x-4=0\) | A1* | Correct proof; brackets must be multiplied out before proceeding to given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^3+6x^2+x-4 = (x+1)(x^2+5x-4)\) | M1 | Divides by \((x+1)\); may be implied by correct quadratic |
| \(x = \frac{-5 \pm \sqrt{25+16}}{2} \Rightarrow x = \frac{-5 \pm \sqrt{41}}{2}\) | dM1 A1 | dM1 for attempting roots of quadratic factor; A1 for \(\frac{-5 \pm \sqrt{41}}{2}\) or exact equivalent; answers with no working score 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = \frac{-5+\sqrt{41}}{2}\) | B1 | Only scorable if \(\frac{-5+\sqrt{41}}{2}\) found in (b)(i); exact answer only (accept awrt 0.702 if given in (b)(i)) |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\log_4(x+3) = \log_4(x+3)^2$ or $\frac{1}{2} = \log_4 2$ | B1 | May be implied by further work |
| Combines two terms e.g. $2\log_4(x+3) + \log_4 x = \log_4 x(x+3)^2$ | M1 | Must correctly combine at least two original terms; $2\log_4(x+3) + \log_4 x = 2\log_4 x(x+3)$ is M0 |
| e.g. $x(x+3)^2 = 2(4x+2)$ | A1 | Correct intermediate equation not involving logs (but not the given answer) |
| e.g. $x(x^2+6x+9) = 8x+4 \Rightarrow x^3+6x^2+x-4=0$ | A1* | Correct proof; brackets must be multiplied out before proceeding to given answer |
## Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^3+6x^2+x-4 = (x+1)(x^2+5x-4)$ | M1 | Divides by $(x+1)$; may be implied by correct quadratic |
| $x = \frac{-5 \pm \sqrt{25+16}}{2} \Rightarrow x = \frac{-5 \pm \sqrt{41}}{2}$ | dM1 A1 | dM1 for attempting roots of quadratic factor; A1 for $\frac{-5 \pm \sqrt{41}}{2}$ or exact equivalent; answers with no working score 0 |
## Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = \frac{-5+\sqrt{41}}{2}$ | B1 | Only scorable if $\frac{-5+\sqrt{41}}{2}$ found in (b)(i); exact answer only (accept awrt 0.702 if given in (b)(i)) |
---\begin{enumerate}
\item (a) Given that
\end{enumerate}
$$2 \log _ { 4 } ( x + 3 ) + \log _ { 4 } x = \log _ { 4 } ( 4 x + 2 ) + \frac { 1 } { 2 }$$
show that
$$x ^ { 3 } + 6 x ^ { 2 } + x - 4 = 0$$
(b) Given also that - 1 is a root of the equation
$$x ^ { 3 } + 6 x ^ { 2 } + x - 4 = 0$$
(i) use algebra to find the other two roots of the equation.\\
(ii) Hence solve
$$2 \log _ { 4 } ( x + 3 ) + \log _ { 4 } x = \log _ { 4 } ( 4 x + 2 ) + \frac { 1 } { 2 }$$
\hfill \mbox{\textit{Edexcel P2 2024 Q6 [8]}}