| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Logarithmic equation solving |
| Difficulty | Moderate -0.3 Part (i) is a straightforward logarithmic equation requiring only taking logs of both sides and dividing (a = ln(70)/ln(3)), which is routine A-level technique. Part (ii) requires applying log laws to collect terms and solve algebraically, which is slightly more involved but still a standard textbook exercise with no novel insight required. The 'show working' requirement and exact answer in (ii) add minor complexity, but overall this is slightly easier than average. |
| 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 |
| \(3^a = 70 \Rightarrow a\log 3 = \log 70\) | M1 | Takes same log of both sides and uses power law; alternatively writes \(a = \log_3 70\). Note: \(a = \log 70\) is M0 |
| \(a = \dfrac{\log 70}{\log 3} = 3.867\) | A1 | CSO; 3.867 following suitable intermediate line; not achieved via Trial and Improvement |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4 = \log_3 81\) or \(3\log_3 b = \log_3 b^3\) or \(\log_3 5b = \log_3 5 + \log_3 b\) | B1 | One correct log law used on a term of the initial equation; can be achieved without bases being present |
| \(4 + 3\log_3 b = \log_3 5b \Rightarrow \log_3 81b^3 = \log_3 5b\) | M1 | Correctly combines two terms of the initial equation to form a log equation of correct form \(\log_3\ldots = \log_3\ldots\) or \(\log_3\ldots = \pm 4\) |
| \(\Rightarrow 81b^2 = 5b \Rightarrow b = \ldots\) | dM1 | Removes logs correctly and proceeds to value for \(b\); must follow M1 |
| \(b = \sqrt{\dfrac{5}{81}}\) | A1 | Or exact equivalent e.g. \(b = \dfrac{1}{9}\sqrt{5}\); negative and zero solutions must be discarded |
# Question 5:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3^a = 70 \Rightarrow a\log 3 = \log 70$ | M1 | Takes same log of both sides and uses power law; alternatively writes $a = \log_3 70$. Note: $a = \log 70$ is M0 |
| $a = \dfrac{\log 70}{\log 3} = 3.867$ | A1 | CSO; 3.867 following suitable intermediate line; not achieved via Trial and Improvement |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4 = \log_3 81$ or $3\log_3 b = \log_3 b^3$ or $\log_3 5b = \log_3 5 + \log_3 b$ | B1 | One correct log law used on a term of the initial equation; can be achieved without bases being present |
| $4 + 3\log_3 b = \log_3 5b \Rightarrow \log_3 81b^3 = \log_3 5b$ | M1 | Correctly combines two terms of the initial equation to form a log equation of correct form $\log_3\ldots = \log_3\ldots$ or $\log_3\ldots = \pm 4$ |
| $\Rightarrow 81b^2 = 5b \Rightarrow b = \ldots$ | dM1 | Removes logs correctly and proceeds to value for $b$; must follow M1 |
| $b = \sqrt{\dfrac{5}{81}}$ | A1 | Or exact equivalent e.g. $b = \dfrac{1}{9}\sqrt{5}$; negative and zero solutions must be discarded |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
Solutions relying entirely on calculator technology are not acceptable.\\
(i) Solve
$$3 ^ { a } = 70$$
giving the answer to 3 decimal places.\\
(ii) Find the exact value of $b$ such that
$$4 + 3 \log _ { 3 } b = \log _ { 3 } 5 b$$
\hfill \mbox{\textit{Edexcel P2 2023 Q5 [6]}}