| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Natural logarithm equation solving |
| Difficulty | Moderate -0.3 Part (a) is routine algebraic factorization and simplification. Part (b) requires applying logarithm laws (difference of logs = log of quotient) and recognizing the connection to part (a), then solving a straightforward rational equation. This is a standard C3 logarithm question with clear structure and no novel insight required, making it slightly easier than average. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(\frac{(x+3)(x+4)}{(2x+1)(x+4)} = \frac{x+3}{2x+1}\) | M1 A2 |
| (b) | \(\ln(x^2 + 7x + 12) - \ln(2x^2 + 9x + 4) = 1, \quad \ln\frac{x^2+7x+12}{2x^2+9x+4} = 1\) | M1 |
| \(\ln\frac{x+3}{2x+1} = 1\) | A1 | |
| \(\frac{x+3}{2x+1} = e\) | A1 | |
| \(x + 3 = e(2x + 1), \quad 3 - e = x(2e - 1)\) | M1 | |
| \(x = \frac{3-e}{2e-1}\) | A1 | |
| (7) |
(a) | $\frac{(x+3)(x+4)}{(2x+1)(x+4)} = \frac{x+3}{2x+1}$ | M1 A2 |
(b) | $\ln(x^2 + 7x + 12) - \ln(2x^2 + 9x + 4) = 1, \quad \ln\frac{x^2+7x+12}{2x^2+9x+4} = 1$ | M1 |
| $\ln\frac{x+3}{2x+1} = 1$ | A1 |
| $\frac{x+3}{2x+1} = e$ | A1 |
| $x + 3 = e(2x + 1), \quad 3 - e = x(2e - 1)$ | M1 |
| $x = \frac{3-e}{2e-1}$ | A1 |
| | (7) |\begin{enumerate}
\item (a) Simplify
\end{enumerate}
$$\frac { x ^ { 2 } + 7 x + 12 } { 2 x ^ { 2 } + 9 x + 4 }$$
(b) Solve the equation
$$\ln \left( x ^ { 2 } + 7 x + 12 \right) - 1 = \ln \left( 2 x ^ { 2 } + 9 x + 4 \right)$$
giving your answer in terms of e.\\
\hfill \mbox{\textit{Edexcel C3 Q1 [7]}}