Question 5 - A-Level Maths

Edexcel C2 2010 January — Question 5 8 marks

Exam Board	Edexcel
Module	C2 (Core Mathematics 2)
Year	2010
Session	January
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Laws of Logarithms
Type	Two unrelated log parts: both solve equations
Difficulty	Moderate -0.3 Part (a) is a straightforward application of the definition of logarithms (x² = 64, so x = 8). Part (b) requires using log laws to simplify (bringing the coefficient down as a power, combining logs) then solving the resulting equation, but follows a standard template for C2 logarithm questions with no unusual complications. Slightly easier than average due to the routine nature of both parts.
Spec	1.06c Logarithm definition: log_a(x) as inverse of a^x 1.06f Laws of logarithms: addition, subtraction, power rules 1.06g Equations with exponentials: solve a^x = b

5. (a) Find the positive value of $x$ such that $$\log _ { x } 64 = 2$$ (b) Solve for $x$ $$\log _ { 2 } ( 11 - 6 x ) = 2 \log _ { 2 } ( x - 1 ) + 3$$

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Question 5:

Part (a):

Answer	Marks	Guidance
$\log_x 64 = 2 \Rightarrow 64 = x^2$, so $x = 8$	M1, A1 (2)	M1 for getting out of logs; $x = 8$ with no working is M1A1; ignore $x = -8$

Part (b):

Answer	Marks	Guidance
$\log_2(11-6x) = \log_2(x-1)^2 + 3$	M1	1st M1 for using the $n\log x$ rule
$\log_2\left[\frac{11-6x}{(x-1)^2}\right] = 3$	M1	2nd M1 for using $\log x - \log y$ or $\log x + \log y$ rule
$\frac{11-6x}{(x-1)^2} = 2^3$	M1	3rd M1 for using 2 to the power; need to see $2^3$ or 8
$11-6x = 8(x^2-2x+1)$, so $0 = 8x^2-10x-3$	A1	1st A1 for correct 3TQ; if all three M marks earned and logs still present, do not give final M1
$0 = (4x+1)(2x-3) \Rightarrow x = \ldots$	dM1	4th dependent M1 for attempt to solve/factorize 3TQ
$x = \frac{3}{2}, \left[-\frac{1}{4}\right]$	A1 (6)	2nd A1 for 1.5 (ignore $-0.25$); s.c. 1.5 only with no working = 0 marks

## Question 5:

### Part (a):
| $\log_x 64 = 2 \Rightarrow 64 = x^2$, so $x = 8$ | M1, A1 (2) | M1 for getting out of logs; $x = 8$ with no working is M1A1; ignore $x = -8$ |

### Part (b):
| $\log_2(11-6x) = \log_2(x-1)^2 + 3$ | M1 | 1st M1 for using the $n\log x$ rule |
| $\log_2\left[\frac{11-6x}{(x-1)^2}\right] = 3$ | M1 | 2nd M1 for using $\log x - \log y$ or $\log x + \log y$ rule |
| $\frac{11-6x}{(x-1)^2} = 2^3$ | M1 | 3rd M1 for using 2 to the power; need to see $2^3$ or 8 |
| $11-6x = 8(x^2-2x+1)$, so $0 = 8x^2-10x-3$ | A1 | 1st A1 for correct 3TQ; if all three M marks earned and logs still present, do not give final M1 |
| $0 = (4x+1)(2x-3) \Rightarrow x = \ldots$ | dM1 | 4th dependent M1 for attempt to solve/factorize 3TQ |
| $x = \frac{3}{2}, \left[-\frac{1}{4}\right]$ | A1 (6) | 2nd A1 for 1.5 (ignore $-0.25$); s.c. 1.5 only with no working = 0 marks |

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5. (a) Find the positive value of $x$ such that

$$\log _ { x } 64 = 2$$

(b) Solve for $x$

$$\log _ { 2 } ( 11 - 6 x ) = 2 \log _ { 2 } ( x - 1 ) + 3$$

\hfill \mbox{\textit{Edexcel C2 2010 Q5 [8]}}

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Q1 4 Q2 6 Q3 9 Q4 7 Q5 8 Q6 9 Q7 10 Q8 12 Q9 10

Answer	Marks	Guidance
\(\log_2(11-6x) = \log_2(x-1)^2 + 3\)	M1	1st M1 for using the \(n\log x\) rule
\(\log_2\left[\frac{11-6x}{(x-1)^2}\right] = 3\)	M1	2nd M1 for using \(\log x - \log y\) or \(\log x + \log y\) rule
\(\frac{11-6x}{(x-1)^2} = 2^3\)	M1	3rd M1 for using 2 to the power; need to see \(2^3\) or 8
\(11-6x = 8(x^2-2x+1)\), so \(0 = 8x^2-10x-3\)	A1	1st A1 for correct 3TQ; if all three M marks earned and logs still present, do not give final M1
\(0 = (4x+1)(2x-3) \Rightarrow x = \ldots\)	dM1	4th dependent M1 for attempt to solve/factorize 3TQ
\(x = \frac{3}{2}, \left[-\frac{1}{4}\right]\)	A1 (6)	2nd A1 for 1.5 (ignore \(-0.25\)); s.c. 1.5 only with no working = 0 marks