| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Logarithmic equation solving |
| Difficulty | Moderate -0.3 Part (a) is a straightforward one-step logarithmic conversion (take log of both sides). Part (b) requires applying log laws to simplify and solve a linear equation, which is standard C2 material but involves more algebraic manipulation than part (a). Both are routine textbook exercises with no problem-solving insight required, making this slightly easier than average overall. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(4^a = 20 \Rightarrow a\log4 = \log20 \Rightarrow a = ...\) | M1 | Takes logs of both sides leading to \(a=...\). Accept \(\log_4 20\), \(\frac{\log20}{\log4}\), or \(a\log4=\log20\Rightarrow a=..\) |
| \(a = \frac{\log20}{\log4} =\) awrt \(2.16\) | A1 | Just the answer with no incorrect working scores both marks. Do not accept trial and error. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3 + 2\log_2 b = \log_2 30b \Rightarrow 3 + \log_2 b^2 = \log_2 30b\) | M1 | Correct use of power law. Accept \(2\log_2 b = \log_2 b^2\), \(2\log b = \log b^2\), or \(3=\log_2 8\). |
| \(\Rightarrow 3 = \log_2 30b - \log_2 b^2\) | M1 | Use of addition or subtraction law of logs. e.g. \(\log(30b)-\log(b^2)=\log\!\left(\frac{30b}{b^2}\right)\). Do not accept \(\log(30b)-2\log(b)=\log\!\left(\frac{30b}{2b}\right)\). |
| \(\Rightarrow 3 = \log_2\!\left(\frac{30b}{b^2}\right)\) | A1 | Correct intermediate line of form \(\log_2(..)=..\) or \(\log_2...=\log_2...\). Accept equivalents such as \(\log_2 8b = \log_2 30\). |
| \(\Rightarrow 2^3 = \frac{30b}{b^2} \Rightarrow b = ..\) | dM1 | Dependent on both previous M marks. Correctly undoing logs and solving for \(b\). |
| \(b = 3.75\) | A1 | \(b=3.75\) or exact equivalent. Ignore any reference to \(b=0\). |
# Question 3:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4^a = 20 \Rightarrow a\log4 = \log20 \Rightarrow a = ...$ | M1 | Takes logs of both sides leading to $a=...$. Accept $\log_4 20$, $\frac{\log20}{\log4}$, or $a\log4=\log20\Rightarrow a=..$ |
| $a = \frac{\log20}{\log4} =$ awrt $2.16$ | A1 | Just the answer with no incorrect working scores both marks. Do not accept trial and error. |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3 + 2\log_2 b = \log_2 30b \Rightarrow 3 + \log_2 b^2 = \log_2 30b$ | M1 | Correct use of power law. Accept $2\log_2 b = \log_2 b^2$, $2\log b = \log b^2$, or $3=\log_2 8$. |
| $\Rightarrow 3 = \log_2 30b - \log_2 b^2$ | M1 | Use of addition or subtraction law of logs. e.g. $\log(30b)-\log(b^2)=\log\!\left(\frac{30b}{b^2}\right)$. Do not accept $\log(30b)-2\log(b)=\log\!\left(\frac{30b}{2b}\right)$. |
| $\Rightarrow 3 = \log_2\!\left(\frac{30b}{b^2}\right)$ | A1 | Correct intermediate line of form $\log_2(..)=..$ or $\log_2...=\log_2...$. Accept equivalents such as $\log_2 8b = \log_2 30$. |
| $\Rightarrow 2^3 = \frac{30b}{b^2} \Rightarrow b = ..$ | dM1 | Dependent on both previous M marks. Correctly undoing logs and solving for $b$. |
| $b = 3.75$ | A1 | $b=3.75$ or exact equivalent. Ignore any reference to $b=0$. |
---3. Solve, giving each answer to 3 significant figures, the equations
\begin{enumerate}[label=(\alph*)]
\item $4 ^ { a } = 20$
\item $3 + 2 \log _ { 2 } b = \log _ { 2 } ( 30 b )$\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2014 Q3 [7]}}