| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Natural logarithm equation solving |
| Difficulty | Moderate -0.3 This is a standard C3 logarithm and exponential question covering routine techniques: solving ln equations by exponentiating, using log laws to combine terms, finding inverse functions, and function composition. All parts follow textbook methods with no novel problem-solving required, making it slightly easier than average but still requiring multiple techniques across several parts. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(e^{\ln(3x-7)} = e^5\) | M1 | Takes \(e\) of both sides; implied by \(3x - 7 = e^5\) |
| \(3x - 7 = e^5 \Rightarrow x = \frac{e^5 + 7}{3}\ \{=51.804\ldots\}\) | dM1, A1 | Exact answer \(\frac{e^5+7}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\ln(3^x e^{7x+2}) = \ln 15\) | M1 | Takes ln of both sides |
| \(\ln 3^x + \ln e^{7x+2} = \ln 15\) | M1 | Applies addition law of logarithms |
| \(x\ln 3 + 7x + 2 = \ln 15\) | A1 oe | |
| \(x(\ln 3 + 7) = -2 + \ln 15\) | ddM1 | Factorising at least two \(x\) terms; collecting number terms |
| \(x = \frac{-2 + \ln 15}{7 + \ln 3}\ \{=0.0874\ldots\}\) | A1 oe | Exact answer \(\frac{-2+\ln 15}{7+\ln 3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = e^{2x} + 3 \Rightarrow y - 3 = e^{2x}\) | M1 | Attempt to make \(x\) (or swapped \(y\)) the subject |
| \(\ln(y-3) = 2x \Rightarrow x = \frac{1}{2}\ln(y-3)\) | M1 | Makes \(e^{2x}\) the subject and takes ln of both sides |
| \(f^{-1}(x) = \frac{1}{2}\ln(x-3)\) | A1 cao | Or \(\frac{1}{2}\ln(x-3)\) or \(\ln\sqrt{x-3}\) |
| Domain: \(x > 3\) or \((3, \infty)\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(fg(x) = e^{2\ln(x-1)} + 3\ \{= (x-1)^2 + 3\}\) | M1, A1 isw | \(e^{2\ln(x-1)} + 3\) or \((x-1)^2 + 3\) or \(x^2 - 2x + 4\) |
| Range of \(fg(x)\): \(y > 3\) or \((3, \infty)\) | B1 | Either \(y>3\) or \((3,\infty)\) or Range \(> 3\) |
## Question 9:
### Part (i)(a): $\ln(3x-7) = 5$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $e^{\ln(3x-7)} = e^5$ | M1 | Takes $e$ of both sides; implied by $3x - 7 = e^5$ |
| $3x - 7 = e^5 \Rightarrow x = \frac{e^5 + 7}{3}\ \{=51.804\ldots\}$ | dM1, A1 | Exact answer $\frac{e^5+7}{3}$ |
### Part (i)(b): $3^x e^{7x+2} = 15$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\ln(3^x e^{7x+2}) = \ln 15$ | M1 | Takes ln of both sides |
| $\ln 3^x + \ln e^{7x+2} = \ln 15$ | M1 | Applies addition law of logarithms |
| $x\ln 3 + 7x + 2 = \ln 15$ | A1 oe | |
| $x(\ln 3 + 7) = -2 + \ln 15$ | ddM1 | Factorising at least two $x$ terms; collecting number terms |
| $x = \frac{-2 + \ln 15}{7 + \ln 3}\ \{=0.0874\ldots\}$ | A1 oe | Exact answer $\frac{-2+\ln 15}{7+\ln 3}$ |
### Part (ii)(a): $f(x) = e^{2x} + 3,\ x \in \mathbb{R}$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = e^{2x} + 3 \Rightarrow y - 3 = e^{2x}$ | M1 | Attempt to make $x$ (or swapped $y$) the subject |
| $\ln(y-3) = 2x \Rightarrow x = \frac{1}{2}\ln(y-3)$ | M1 | Makes $e^{2x}$ the subject and takes ln of both sides |
| $f^{-1}(x) = \frac{1}{2}\ln(x-3)$ | A1 cao | Or $\frac{1}{2}\ln(x-3)$ or $\ln\sqrt{x-3}$ |
| Domain: $x > 3$ or $(3, \infty)$ | B1 | |
### Part (ii)(b): $g(x) = \ln(x-1),\ x \in \mathbb{R},\ x > 1$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $fg(x) = e^{2\ln(x-1)} + 3\ \{= (x-1)^2 + 3\}$ | M1, A1 isw | $e^{2\ln(x-1)} + 3$ or $(x-1)^2 + 3$ or $x^2 - 2x + 4$ |
| Range of $fg(x)$: $y > 3$ or $(3, \infty)$ | B1 | Either $y>3$ or $(3,\infty)$ or Range $> 3$ |
The image provided appears to be a **back/end page** of an Edexcel publication — it contains only publisher contact information (address, telephone, fax, email, order code) and Edexcel's registered company details.
**There is no mark scheme content on this page** — no questions, answers, mark allocations, or guidance notes to extract.
If you have additional pages from the mark scheme, please share those and I'll be happy to extract and format the content for you.9. (i) Find the exact solutions to the equations
\begin{enumerate}[label=(\alph*)]
\item $\ln ( 3 x - 7 ) = 5$
\item $3 ^ { x } \mathrm { e } ^ { 7 x + 2 } = 15$\\
(ii) The functions f and g are defined by
$$\begin{array} { l l }
\mathrm { f } ( x ) = \mathrm { e } ^ { 2 x } + 3 , & x \in \mathbb { R } \\
\mathrm {~g} ( x ) = \ln ( x - 1 ) , & x \in \mathbb { R } , x > 1
\end{array}$$
(a) Find $\mathrm { f } ^ { - 1 }$ and state its domain.\\
(b) Find fg and state its range.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2010 Q9 [15]}}