Question 6 - A-Level Maths

Edexcel P2 2018 Specimen — Question 6 7 marks

Exam Board	Edexcel
Module	P2 (Pure Mathematics 2)
Year	2018
Session	Specimen
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Laws of Logarithms
Type	Two unrelated log parts: both solve equations
Difficulty	Moderate -0.8 This question tests routine application of logarithm laws with straightforward algebraic manipulation. Part (i) requires using the subtraction law and solving a linear equation; part (ii) involves the power law and rearranging. Both are standard textbook exercises with no problem-solving insight required, making this easier than average.
Spec	1.06f Laws of logarithms: addition, subtraction, power rules

6. (i) Find the exact value of $x$ for which $$\log _ { 2 } ( 2 x ) = \log _ { 2 } ( 5 x + 4 ) - 3$$ (ii) Given that $$\log _ { a } y + 3 \log _ { a } 2 = 5$$ express $y$ in terms of $a$. Give your answer in its simplest form. \includegraphics[max width=\textwidth, alt={}, center]{0aafa21b-25f4-4f36-b914-bbaf6cae7a66-18_2674_1948_107_118}

Show mark scheme Show mark scheme source

Question 6:

Part (i):

Answer	Marks	Guidance
$\log_2\!\left(\frac{2x}{5x+4}\right) = -3$ or $\log_2\!\left(\frac{5x+4}{2x}\right) = 3$ or $\log_2\!\left(\frac{5x+4}{x}\right) = 4$	M1	Applying subtraction or addition law of logs correctly to make two log terms into one
$\left(\frac{2x}{5x+4}\right) = 2^{-3}$ or $\left(\frac{5x+4}{2x}\right) = 2^3$ or $\left(\frac{5x+4}{x}\right) = 2^4$	M1	For RHS of either $2^{-3}$, $2^3$, $2^4$ or $\log_2\!\left(\frac{1}{8}\right)$, $\log_2 8$ or $\log_2 16$
$16x = 5x + 4 \Rightarrow x = \ldots$	dM1	Obtains correct linear equation in $x$
$x = \frac{4}{11}$ or exact recurring decimal $0.\dot{3}\dot{6}$ after correct work	A1 cso

Alternative:

Answer	Marks	Guidance
$\log_2(2x) + 3 = \log_2(5x+4)$
$\log_2(2x) + \log_2(8) = \log_2(5x+4)$	2nd M1	3 replaced by $\log_2 8$
$\log_2(16x) = \log_2(5x+4)$	1st M1	Addition law of logs
Then final M1 A1 as before	dM1 A1

Part (ii):

Answer	Marks	Guidance
$\log_a y + \log_a 2^3 = 5$	M1	Applies power law of logarithms to replace $3\log_a 2$ by $\log_a 2^3$ or $\log_a 8$
$\log_a 8y = 5$	dM1	Applies product law of logarithms (should not follow M0)
$y = \frac{1}{8}a^5$ cso	A1

## Question 6:

### Part (i):
| $\log_2\!\left(\frac{2x}{5x+4}\right) = -3$ **or** $\log_2\!\left(\frac{5x+4}{2x}\right) = 3$ **or** $\log_2\!\left(\frac{5x+4}{x}\right) = 4$ | M1 | Applying subtraction or addition law of logs correctly to make two log terms into one |
|---|---|---|
| $\left(\frac{2x}{5x+4}\right) = 2^{-3}$ **or** $\left(\frac{5x+4}{2x}\right) = 2^3$ **or** $\left(\frac{5x+4}{x}\right) = 2^4$ | M1 | For RHS of either $2^{-3}$, $2^3$, $2^4$ or $\log_2\!\left(\frac{1}{8}\right)$, $\log_2 8$ or $\log_2 16$ |
| $16x = 5x + 4 \Rightarrow x = \ldots$ | dM1 | Obtains correct linear equation in $x$ |
| $x = \frac{4}{11}$ or exact recurring decimal $0.\dot{3}\dot{6}$ after correct work | A1 cso | |

**Alternative:**
| $\log_2(2x) + 3 = \log_2(5x+4)$ | | |
|---|---|---|
| $\log_2(2x) + \log_2(8) = \log_2(5x+4)$ | 2nd M1 | 3 replaced by $\log_2 8$ |
| $\log_2(16x) = \log_2(5x+4)$ | 1st M1 | Addition law of logs |
| Then final M1 A1 as before | dM1 A1 | |

### Part (ii):
| $\log_a y + \log_a 2^3 = 5$ | M1 | Applies power law of logarithms to replace $3\log_a 2$ by $\log_a 2^3$ or $\log_a 8$ |
|---|---|---|
| $\log_a 8y = 5$ | dM1 | Applies product law of logarithms (should not follow M0) |
| $y = \frac{1}{8}a^5$ **cso** | A1 | |

Show LaTeX source

6. (i) Find the exact value of $x$ for which

$$\log _ { 2 } ( 2 x ) = \log _ { 2 } ( 5 x + 4 ) - 3$$

(ii) Given that

$$\log _ { a } y + 3 \log _ { a } 2 = 5$$

express $y$ in terms of $a$. Give your answer in its simplest form.\\

\includegraphics[max width=\textwidth, alt={}, center]{0aafa21b-25f4-4f36-b914-bbaf6cae7a66-18_2674_1948_107_118}\\

\hfill \mbox{\textit{Edexcel P2 2018 Q6 [7]}}

This paper (9 questions)

View full paper

Q1 7 Q2 8 Q3 7 Q4 4 Q5 11 Q6 7 Q7 10 Q8 12 Q9 9

Answer	Marks	Guidance
\(\log_2\!\left(\frac{2x}{5x+4}\right) = -3\) or \(\log_2\!\left(\frac{5x+4}{2x}\right) = 3\) or \(\log_2\!\left(\frac{5x+4}{x}\right) = 4\)	M1	Applying subtraction or addition law of logs correctly to make two log terms into one
\(\left(\frac{2x}{5x+4}\right) = 2^{-3}\) or \(\left(\frac{5x+4}{2x}\right) = 2^3\) or \(\left(\frac{5x+4}{x}\right) = 2^4\)	M1	For RHS of either \(2^{-3}\), \(2^3\), \(2^4\) or \(\log_2\!\left(\frac{1}{8}\right)\), \(\log_2 8\) or \(\log_2 16\)
\(16x = 5x + 4 \Rightarrow x = \ldots\)	dM1	Obtains correct linear equation in \(x\)
\(x = \frac{4}{11}\) or exact recurring decimal \(0.\dot{3}\dot{6}\) after correct work	A1 cso

Answer	Marks	Guidance
\(\log_2(2x) + 3 = \log_2(5x+4)\)
\(\log_2(2x) + \log_2(8) = \log_2(5x+4)\)	2nd M1	3 replaced by \(\log_2 8\)
\(\log_2(16x) = \log_2(5x+4)\)	1st M1	Addition law of logs
Then final M1 A1 as before	dM1 A1

Answer	Marks	Guidance
\(\log_a y + \log_a 2^3 = 5\)	M1	Applies power law of logarithms to replace \(3\log_a 2\) by \(\log_a 2^3\) or \(\log_a 8\)
\(\log_a 8y = 5\)	dM1	Applies product law of logarithms (should not follow M0)
\(y = \frac{1}{8}a^5\) cso	A1