| 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 | Standard +0.3 This question requires knowledge of logarithm change of base formula and algebraic manipulation to convert to a common variable, then solving a linear equation in y. While it involves multiple steps and logarithm properties, the techniques are standard C3 material with clear scaffolding in part (a) that guides students to the solution method. |
| Spec | 1.06c Logarithm definition: log_a(x) as inverse of a^x1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| (a) (i) \(= \frac{\ln x}{\ln 2} = \frac{y}{\ln 2}\) | M1 A1 | |
| (ii) \(= \ln x^2 - \ln e = 2 \ln x - 1 = 2y - 1\) | M1 A1 | |
| (b) \(\frac{y}{\ln 2} = 4 - (2y - 1)\) | M1 | |
| \(y = (5 - 2y)\ln 2\) | M1 | |
| \(y(2 \ln 2 + 1) = 5 \ln 2\) | M1 | |
| \(y = \frac{5\ln 2}{2\ln 2 + 1}\) | A1 | |
| \(x = e^4 = 4.27\) (2dp) | A1 | (8 marks) |
**(a)** (i) $= \frac{\ln x}{\ln 2} = \frac{y}{\ln 2}$ | M1 A1
(ii) $= \ln x^2 - \ln e = 2 \ln x - 1 = 2y - 1$ | M1 A1
**(b)** $\frac{y}{\ln 2} = 4 - (2y - 1)$ | M1
$y = (5 - 2y)\ln 2$ | M1
$y(2 \ln 2 + 1) = 5 \ln 2$ | M1
$y = \frac{5\ln 2}{2\ln 2 + 1}$ | A1
$x = e^4 = 4.27$ (2dp) | A1 | (8 marks)3. (a) Given that $y = \ln x$, find expressions in terms of $y$ for
\begin{enumerate}[label=(\roman*)]
\item $\quad \log _ { 2 } x$,
\item $\ln \frac { x ^ { 2 } } { \mathrm { e } }$.\\
(b) Hence, or otherwise, solve the equation
$$\log _ { 2 } x = 4 - \ln \frac { x ^ { 2 } } { \mathrm { e } } ,$$
giving your answer to 2 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q3 [8]}}