Question 7 - A-Level Maths

Edexcel P2 2021 June — Question 7 10 marks

Exam Board	Edexcel
Module	P2 (Pure Mathematics 2)
Year	2021
Session	June
Marks	10
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 structured multi-part question with clear signposting that guides students through each step: applying log laws to reach a polynomial, verifying a root, then factoring to solve. While it requires multiple techniques (log laws, polynomial manipulation, factor theorem), the scaffolding makes it slightly easier than average, and each individual step is routine for P2 level.
Spec	1.02j Manipulate polynomials: expanding, factorising, division, factor theorem 1.06f Laws of logarithms: addition, subtraction, power rules

7. (a) Given that $$3 \log _ { 3 } ( 2 x - 1 ) = 2 + \log _ { 3 } ( 14 x - 25 )$$ show that $$2 x ^ { 3 } - 3 x ^ { 2 } - 30 x + 56 = 0$$ (b) Show that - 4 is a root of this cubic equation.
(c) Hence, using algebra and showing each step of your working, solve $$3 \log _ { 3 } ( 2 x - 1 ) = 2 + \log _ { 3 } ( 14 x - 25 )$$

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Question 7:

Part (a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$2 = \log_3 3^2$ (seen or implied)	B1	May be gained via $\log_3 f(x)=2 \Rightarrow f(x)=9$
$3\log_3(2x-1) = \log_3(2x-1)^3$ or $\log_3\text{"3}^2\text{"} + \log_3(14x-25) = \log_3\left(\text{"3}^2\text{"}\times(14x-25)\right)$	M1	Uses correct log law (power or sum)
$\Rightarrow (2x-1)^3 = 9(14x-25)$	dM1	Uses correct log work to remove logs; all log work must be correct
$\Rightarrow 8x^3 - 12x^2 + 6x - 1 = 126x - 225$ $\Rightarrow 8x^3 - 12x^2 - 120x + 224 = 0 \Rightarrow 2x^3 - 3x^2 - 30x + 56 = 0$	A1*	Expands and simplifies to given cubic; must show intermediate step

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$2(\pm4)^3 - 3(\pm4)^2 - 30(\pm4) + 56 = \ldots$	M1
$2(-4)^3 - 3(-4)^2 - 30(-4) + 56 = -128 - 48 + 120 + 56 = 0$; hence $-4$ is a root	A1

Part (b) — Alternative:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$2x^3 - 3x^2 - 30x + 56 = (x\pm4)(2x^2 + \ldots \pm14)$	M1
$2x^3 - 3x^2 - 30x + 56 = (x+4)(2x^2-11x+14)$, so $-4$ is a root	A1

Part (c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$2x^3 - 3x^2 - 30x + 56 = 0 \Rightarrow (x+4)(2x^2 + \ldots \pm14) = 0$	M1
$(x+4)(2x^2 - 11x + 14) = 0$	A1
$(x+4)(2x-7)(x-2) = 0 \Rightarrow x = \ldots$	dM1
Equation not defined for $x=-4$, so solutions are $2$ and $\frac{7}{2}$	A1

## Question 7:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $2 = \log_3 3^2$ (seen or implied) | B1 | May be gained via $\log_3 f(x)=2 \Rightarrow f(x)=9$ |
| $3\log_3(2x-1) = \log_3(2x-1)^3$ or $\log_3\text{"3}^2\text{"} + \log_3(14x-25) = \log_3\left(\text{"3}^2\text{"}\times(14x-25)\right)$ | M1 | Uses correct log law (power or sum) |
| $\Rightarrow (2x-1)^3 = 9(14x-25)$ | dM1 | Uses correct log work to remove logs; all log work must be correct |
| $\Rightarrow 8x^3 - 12x^2 + 6x - 1 = 126x - 225$ $\Rightarrow 8x^3 - 12x^2 - 120x + 224 = 0 \Rightarrow 2x^3 - 3x^2 - 30x + 56 = 0$ | A1* | Expands and simplifies to given cubic; must show intermediate step |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $2(\pm4)^3 - 3(\pm4)^2 - 30(\pm4) + 56 = \ldots$ | M1 | |
| $2(-4)^3 - 3(-4)^2 - 30(-4) + 56 = -128 - 48 + 120 + 56 = 0$; hence $-4$ is a root | A1 | |

### Part (b) — Alternative:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $2x^3 - 3x^2 - 30x + 56 = (x\pm4)(2x^2 + \ldots \pm14)$ | M1 | |
| $2x^3 - 3x^2 - 30x + 56 = (x+4)(2x^2-11x+14)$, so $-4$ is a root | A1 | |

### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $2x^3 - 3x^2 - 30x + 56 = 0 \Rightarrow (x+4)(2x^2 + \ldots \pm14) = 0$ | M1 | |
| $(x+4)(2x^2 - 11x + 14) = 0$ | A1 | |
| $(x+4)(2x-7)(x-2) = 0 \Rightarrow x = \ldots$ | dM1 | |
| Equation not defined for $x=-4$, so solutions are $2$ and $\frac{7}{2}$ | A1 | |

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7. (a) Given that

$$3 \log _ { 3 } ( 2 x - 1 ) = 2 + \log _ { 3 } ( 14 x - 25 )$$

show that

$$2 x ^ { 3 } - 3 x ^ { 2 } - 30 x + 56 = 0$$

(b) Show that - 4 is a root of this cubic equation.\\
(c) Hence, using algebra and showing each step of your working, solve

$$3 \log _ { 3 } ( 2 x - 1 ) = 2 + \log _ { 3 } ( 14 x - 25 )$$

\hfill \mbox{\textit{Edexcel P2 2021 Q7 [10]}}

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