Question 6 - A-Level Maths

Edexcel C2 2007 June — Question 6 6 marks

Exam Board	Edexcel
Module	C2 (Core Mathematics 2)
Year	2007
Session	June
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Exponential Equations & Modelling
Type	Logarithmic equation solving
Difficulty	Moderate -0.3 Part (a) is a straightforward exponential equation requiring taking logs of both sides—a standard C2 technique. Part (b) involves applying log laws to simplify and solve, which is routine but requires careful algebraic manipulation. Both parts are standard textbook exercises with no novel insight required, making this slightly easier than average.
Spec	1.06f Laws of logarithms: addition, subtraction, power rules 1.06g Equations with exponentials: solve a^x = b

6. (a) Find, to 3 significant figures, the value of $x$ for which $8 ^ { x } = 0.8$.
(b) Solve the equation $$2 \log _ { 3 } x - \log _ { 3 } 7 x = 1$$

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Question 6:

Part (a):

Answer	Marks	Guidance
$x = \frac{\log 0.8}{\log 8}$ or $\log_8 0.8$	M1, A1	Allow awrt $-0.107$; allow 'implicit' answer $8^{-0.107}$; answer $-0.11$ or awrt scores M1 A0

Part (b):

Answer	Marks	Guidance
$2\log x = \log x^2$	B1
$\log x^2 - \log 7x = \log \frac{x^2}{7x}$	M1
"Remove logs" to form equation in $x$: $\frac{x^2}{7x} = 3$	M1	Using base correctly
$x = 21$	A1cso	Ignore $x = 0$ if seen

Alternative:

Answer	Marks	Guidance
$\log 7x + 1 = \log 7x + \log 3 = \log 21x$	M1
$x^2 = 21x$	M1
$x = 21$	A1	Ignore $x = 0$ if seen

Alternative:

Answer	Marks
$\log 7x = \log 7 + \log x$	B1
$\log_3 x = 1 + \log_3 7$	M1
$x = 3^{(1+\log_3 7)}$ or $\log_3 x = \log_3 3 + \log_3 7$	M1
$x = 21$	A1

## Question 6:

### Part (a):
| $x = \frac{\log 0.8}{\log 8}$ or $\log_8 0.8$ | M1, A1 | Allow awrt $-0.107$; allow 'implicit' answer $8^{-0.107}$; answer $-0.11$ or awrt scores M1 A0 |

### Part (b):
| $2\log x = \log x^2$ | B1 | |
| $\log x^2 - \log 7x = \log \frac{x^2}{7x}$ | M1 | |
| "Remove logs" to form equation in $x$: $\frac{x^2}{7x} = 3$ | M1 | Using base correctly |
| $x = 21$ | A1cso | Ignore $x = 0$ if seen |

**Alternative:**
| $\log 7x + 1 = \log 7x + \log 3 = \log 21x$ | M1 | |
| $x^2 = 21x$ | M1 | |
| $x = 21$ | A1 | Ignore $x = 0$ if seen |

**Alternative:**
| $\log 7x = \log 7 + \log x$ | B1 | |
| $\log_3 x = 1 + \log_3 7$ | M1 | |
| $x = 3^{(1+\log_3 7)}$ or $\log_3 x = \log_3 3 + \log_3 7$ | M1 | |
| $x = 21$ | A1 | |

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6. (a) Find, to 3 significant figures, the value of $x$ for which $8 ^ { x } = 0.8$.\\
(b) Solve the equation

$$2 \log _ { 3 } x - \log _ { 3 } 7 x = 1$$

\hfill \mbox{\textit{Edexcel C2 2007 Q6 [6]}}

This paper (10 questions)

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Q1 4 Q2 6 Q3 6 Q4 5 Q5 9 Q6 6 Q7 9 Q8 9 Q9 10 Q10 11

Answer	Marks	Guidance
\(2\log x = \log x^2\)	B1
\(\log x^2 - \log 7x = \log \frac{x^2}{7x}\)	M1
"Remove logs" to form equation in \(x\): \(\frac{x^2}{7x} = 3\)	M1	Using base correctly
\(x = 21\)	A1cso	Ignore \(x = 0\) if seen

Answer	Marks	Guidance
\(\log 7x + 1 = \log 7x + \log 3 = \log 21x\)	M1
\(x^2 = 21x\)	M1
\(x = 21\)	A1	Ignore \(x = 0\) if seen

Answer	Marks
\(\log 7x = \log 7 + \log x\)	B1
\(\log_3 x = 1 + \log_3 7\)	M1
\(x = 3^{(1+\log_3 7)}\) or \(\log_3 x = \log_3 3 + \log_3 7\)	M1
\(x = 21\)	A1