| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Solve log equation reducing to quadratic |
| Difficulty | Moderate -0.3 This is a straightforward application of logarithm laws (power rule, subtraction rule) followed by solving a quadratic equation. The steps are routine: combine logs using laws, convert to exponential form, solve the resulting quadratic. Slightly easier than average as it follows a standard template with no conceptual surprises. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses one correct log law, e.g. \(2\log_5(3x-2) \to \log_5(3x-2)^2\) or \(2 \to \log_5 25\) or \(\log_5 \ldots = 2 \to \ldots = 5^2\) | B1 | 1.1a — Base need not be seen |
| Uses two correct log laws leading to equation without logs, e.g. \(2\log_5(3x-2) - \log_5 x \to \log_5\frac{(3x-2)^2}{x}\) | M1 | 3.1a — Must not follow from incorrect log work |
| \(\frac{(3x-2)^2}{x} = 25\) | A1 | 1.1b — No incorrect work seen |
| \(\frac{(3x-2)^2}{x} = 25 \Rightarrow 9x^2 - 37x + 4 = 0 \Rightarrow (9x-1)(x-4) = 0\) | dM1 | 1.1b — Correct method to solve via 3TQ \(= 0\) |
| \(x = 4\) only | A1 cso | 3.2a — \(x = \frac{1}{9}\) must be rejected; \(x = 0\) must also be rejected if seen |
## Question 9:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses one correct log law, e.g. $2\log_5(3x-2) \to \log_5(3x-2)^2$ **or** $2 \to \log_5 25$ **or** $\log_5 \ldots = 2 \to \ldots = 5^2$ | B1 | 1.1a — Base need not be seen |
| Uses two correct log laws leading to equation without logs, e.g. $2\log_5(3x-2) - \log_5 x \to \log_5\frac{(3x-2)^2}{x}$ | M1 | 3.1a — Must not follow from incorrect log work |
| $\frac{(3x-2)^2}{x} = 25$ | A1 | 1.1b — No incorrect work seen |
| $\frac{(3x-2)^2}{x} = 25 \Rightarrow 9x^2 - 37x + 4 = 0 \Rightarrow (9x-1)(x-4) = 0$ | dM1 | 1.1b — Correct method to solve via 3TQ $= 0$ |
| $x = 4$ only | A1 cso | 3.2a — $x = \frac{1}{9}$ must be rejected; $x = 0$ must also be rejected if seen |
---\begin{enumerate}
\item Using the laws of logarithms, solve the equation
\end{enumerate}
$$2 \log _ { 5 } ( 3 x - 2 ) - \log _ { 5 } x = 2$$
\hfill \mbox{\textit{Edexcel AS Paper 1 2023 Q9 [5]}}