Question 3 - A-Level Maths

Edexcel C3 — Question 3 8 marks

Exam Board	Edexcel
Module	C3 (Core Mathematics 3)
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Exponential Equations & Modelling
Type	Natural logarithm equation solving
Difficulty	Challenging +1.2 This simultaneous equation problem requires students to manipulate exponential and logarithmic expressions strategically, substitute between equations, and solve a resulting quadratic. While it involves multiple C3 techniques (exponentials, logarithms, substitution), the path forward is relatively clear once students recognize they can express one variable in terms of the other. The algebraic manipulation is moderately demanding but follows standard patterns, placing it above average difficulty but not requiring exceptional insight.
Spec	1.02c Simultaneous equations: two variables by elimination and substitution 1.06a Exponential function: a^x and e^x graphs and properties 1.06d Natural logarithm: ln(x) function and properties

3. Giving your answers to 2 decimal places, solve the simultaneous equations $$\begin{aligned} & \mathrm { e } ^ { 2 y } - x + 2 = 0 \\ & \ln ( x + 3 ) - 2 y - 1 = 0 \end{aligned}$$

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Answer	Marks	Guidance
$e^{2y} - x + 2 = 0 \Rightarrow e^{2y} = x - 2$	M1
$2y = \ln(x - 2)$	A1
Sub. $\Rightarrow \ln(x + 3) - \ln(x - 2) - 1 = 0$	M1 A1
$\ln\frac{x + 3}{x - 2} = 1$	M1
$\frac{x + 3}{x - 2} = e$	A1
$x + 3 = e(x - 2)$	M1
$3 + 2e = x(e - 1)$	M1
$x = \frac{2e + 3}{e - 1} = 4.91$ (2dp), $y = \frac{1}{2}\ln\left(\frac{2e + 3}{e - 1} - 2\right) = 0.53$ (2dp)	A2	(8 marks)

$e^{2y} - x + 2 = 0 \Rightarrow e^{2y} = x - 2$ | M1 |

$2y = \ln(x - 2)$ | A1 |

Sub. $\Rightarrow \ln(x + 3) - \ln(x - 2) - 1 = 0$ | M1 A1 |

$\ln\frac{x + 3}{x - 2} = 1$ | M1 |

$\frac{x + 3}{x - 2} = e$ | A1 |

$x + 3 = e(x - 2)$ | M1 |

$3 + 2e = x(e - 1)$ | M1 |

$x = \frac{2e + 3}{e - 1} = 4.91$ (2dp), $y = \frac{1}{2}\ln\left(\frac{2e + 3}{e - 1} - 2\right) = 0.53$ (2dp) | A2 | (8 marks)

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3. Giving your answers to 2 decimal places, solve the simultaneous equations

$$\begin{aligned}
& \mathrm { e } ^ { 2 y } - x + 2 = 0 \\
& \ln ( x + 3 ) - 2 y - 1 = 0
\end{aligned}$$

\hfill \mbox{\textit{Edexcel C3  Q3 [8]}}

This paper (8 questions)

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Q1 4 Q2 6 Q3 8 Q4 10 Q5 10 Q6 12 Q7 12 Q8 13

Answer	Marks	Guidance
\(e^{2y} - x + 2 = 0 \Rightarrow e^{2y} = x - 2\)	M1
\(2y = \ln(x - 2)\)	A1
Sub. \(\Rightarrow \ln(x + 3) - \ln(x - 2) - 1 = 0\)	M1 A1
\(\ln\frac{x + 3}{x - 2} = 1\)	M1
\(\frac{x + 3}{x - 2} = e\)	A1
\(x + 3 = e(x - 2)\)	M1
\(3 + 2e = x(e - 1)\)	M1
\(x = \frac{2e + 3}{e - 1} = 4.91\) (2dp), \(y = \frac{1}{2}\ln\left(\frac{2e + 3}{e - 1} - 2\right) = 0.53\) (2dp)	A2	(8 marks)