| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Logarithmic equation solving |
| Difficulty | Moderate -0.3 Part (i) is a straightforward exponential equation requiring taking logarithms of both sides—a standard C2 technique. Part (ii) involves routine logarithm laws to form a quadratic equation (the 'show that' guides students), then solving it. While multi-step, all techniques are standard C2 content with no novel insight required, making it 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 |
|---|---|---|
| Part | Answer/Working | Marks |
| (i) | Use of power rule so \((y-1)\log 1.01 = \log 500\) or \((y-1) = \log_{101} 500\) | M1 |
| A1 | A1: Accept answers which round to 625.56 (This may follow 624.56 + 1 = or may follow \(y = \log_{1.01} 505\) or may appear with no working) | |
| 625.56 | ||
| (2) | ||
| (ii)(a) | Ignore labels (a) and (b) in part ii and mark work as seen. \(\log_4(3x+5)^2 = \) (Applies power law of logarithms) | M1 |
| Uses \(\log_4 4 = 1\) or \(4^1 = 4\) | M1 | M1: Uses \(\log_4 4 = 1\) or \(4^1 = 4\) |
| Uses quotient or product rule so e.g. \(\log(3x+5)^2 = \log 4(3x+8)\) or \(\log \frac{(3x+5)^2}{(3x+8)}\) | M1 | M1: Applying subtraction or addition law of logarithms correctly to make two log terms into one log term in \(x\) (*see note below*) |
| Obtains with no errors \(9x^2 + 18x - 7 = 0\)* | A1* cso | A1* cso: This is a given answer and needs a correct algebraic statement such as \(9x^2 + 30x + 25 = 4(3x + 8)\) followed by a conclusion, such as \(9x^2 + 18x - 7 = 0\) |
| (4) | ||
| (b) | Solves given or "their" quadratic equation by any of standard methods | M1 |
| Obtains \(x = \frac{1}{3}\) and \(-\frac{7}{3}\) and rejects \(-\frac{7}{3}\) to give just \(\frac{1}{3}\) | A1 | A1: Needs to find two answers and reject one to give correct \(\frac{1}{3}\) (This may be indicated by underlining just 1/3 for example) |
| (2) | ||
| [8] |
| **Part** | **Answer/Working** | **Marks** | **Notes** |
|---|---|---|---|
| (i) | Use of power rule so $(y-1)\log 1.01 = \log 500$ or $(y-1) = \log_{101} 500$ | M1 | M1: Applies power law of logarithms correctly or changes base (Allow missing brackets) |
| | | A1 | A1: Accept answers which round to 625.56 (This may follow 624.56 + 1 = or may follow $y = \log_{1.01} 505$ or may appear with no working) |
| | 625.56 | | |
| | | **(2)** | |
| (ii)(a) | Ignore labels (a) and (b) in part ii and mark work as seen. $\log_4(3x+5)^2 = $ (Applies power law of logarithms) | M1 | M1: Applies power law of logarithms correctly or changes base (Allow missing brackets) |
| | Uses $\log_4 4 = 1$ or $4^1 = 4$ | M1 | M1: Uses $\log_4 4 = 1$ or $4^1 = 4$ |
| | Uses quotient or product rule so e.g. $\log(3x+5)^2 = \log 4(3x+8)$ or $\log \frac{(3x+5)^2}{(3x+8)}$ | M1 | M1: Applying subtraction or addition law of logarithms **correctly** to make two log terms into one log term in $x$ (*see note below*) |
| | Obtains with no errors $9x^2 + 18x - 7 = 0$* | A1* cso | A1* cso: This is a given answer and needs a correct algebraic statement such as $9x^2 + 30x + 25 = 4(3x + 8)$ followed by a conclusion, such as $9x^2 + 18x - 7 = 0$ |
| | | **(4)** | |
| (b) | Solves given or "their" quadratic equation by any of standard methods | M1 | M1: Solves by factorisation or by completion of the square or by correct use of formula (see general principles) |
| | Obtains $x = \frac{1}{3}$ and $-\frac{7}{3}$ and rejects $-\frac{7}{3}$ to give just $\frac{1}{3}$ | A1 | A1: Needs to find two answers and reject one to give correct $\frac{1}{3}$ (This may be indicated by underlining just 1/3 for example) |
| | | **(2)** | |
| | | **[8]** | |
**Notes:**
(i) M1: Applies power law of logarithms correctly or changes base (Allow missing brackets). A1: Accept answers which round to 625.56 (7. (i) Find the value of $y$ for which
$$1.01 ^ { y - 1 } = 500$$
Give your answer to 2 decimal places.\\
(ii) Given that
$$2 \log _ { 4 } ( 3 x + 5 ) = \log _ { 4 } ( 3 x + 8 ) + 1 , \quad x > - \frac { 5 } { 3 }$$
\begin{enumerate}[label=(\alph*)]
\item show that
$$9 x ^ { 2 } + 18 x - 7 = 0$$
\item Hence solve the equation
$$2 \log _ { 4 } ( 3 x + 5 ) = \log _ { 4 } ( 3 x + 8 ) + 1 , \quad x > - \frac { 5 } { 3 }$$
DO NOTI WRITE IN THIS AREA
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2018 Q7 [8]}}