| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Solve ln equation using subtraction law |
| Difficulty | Standard +0.3 This is a straightforward logarithm manipulation question requiring application of log laws (difference rule) and basic algebraic rearrangement. Part (a) involves 3-4 standard steps with no conceptual difficulty, while part (b) tests understanding of domain restrictions. Slightly easier than average due to the guided nature and routine techniques. |
| 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 |
| States \(\log a - \log b = \log\frac{a}{b}\) | B1 | May be implied by starting line \(\frac{a}{b} = a - b\) |
| Proceeds from \(\frac{a}{b} = a - b \rightarrow \ldots \rightarrow ab - a = b^2\) | M1 | Attempts to make \(a\) the subject; two \(a\) terms on same side |
| \(ab - a = b^2 \rightarrow a(b-1) = b^2 \Rightarrow a = \frac{b^2}{b-1}\) * | A1* | Clear reasoning; bracketing must be correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States either \(b > 1\) or \(b \neq 1\) with reason \(\frac{b^2}{b-1}\) is not defined at \(b = 1\) | B1 | May state \(b\) cannot be less than 1 |
| States \(b > 1\) and explains that \(a > 0 \Rightarrow \frac{b^2}{b-1} > 0 \Rightarrow b > 1\) | B1 | As \(b^2\) is positive; accept \(b > 1\) as \(a\) cannot be negative |
## Question 9:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States $\log a - \log b = \log\frac{a}{b}$ | B1 | May be implied by starting line $\frac{a}{b} = a - b$ |
| Proceeds from $\frac{a}{b} = a - b \rightarrow \ldots \rightarrow ab - a = b^2$ | M1 | Attempts to make $a$ the subject; two $a$ terms on same side |
| $ab - a = b^2 \rightarrow a(b-1) = b^2 \Rightarrow a = \frac{b^2}{b-1}$ * | A1* | Clear reasoning; bracketing must be correct |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States either $b > 1$ or $b \neq 1$ with reason $\frac{b^2}{b-1}$ is not defined at $b = 1$ | B1 | May state $b$ cannot be less than 1 |
| States $b > 1$ and explains that $a > 0 \Rightarrow \frac{b^2}{b-1} > 0 \Rightarrow b > 1$ | B1 | As $b^2$ is positive; accept $b > 1$ as $a$ cannot be negative |
---\begin{enumerate}
\item Given that $a > b > 0$ and that $a$ and $b$ satisfy the equation
\end{enumerate}
$$\log a - \log b = \log ( a - b )$$
(a) show that
$$a = \frac { b ^ { 2 } } { b - 1 }$$
(3)\\
(b) Write down the full restriction on the value of $b$, explaining the reason for this restriction.
\hfill \mbox{\textit{Edexcel Paper 1 2019 Q9 [5]}}