| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Evaluate log expression using laws |
| Difficulty | Moderate -0.8 Part (i) is a straightforward application of logarithm laws (sum/difference rules) requiring only direct recall and basic arithmetic. Part (ii) involves rearranging an exponential equation using logarithms, which is a standard textbook exercise with clear steps. Both parts are routine C2-level questions with no problem-solving insight required, making this easier than average. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\log_5\left(\frac{15 \times 20}{12}\right) = \log_5 25 = 2\) | M1, A1, A1 | For any relevant combination of \(\log a \pm \log b\); For \(\log 25\) – must follow correct working only; For correct answer 2 |
| (ii) Method A: \(\frac{1}{3}y = 10^{2x}\) | M1, M1 | For correct division of both sides by 3; For relevant use of \(a = b^c \Rightarrow c = \log_a b\) |
| Hence \(2x = \log_{10}\left(\frac{1}{3}y\right)\) | A1 | For correct equation involving logs to base 10 |
| i.e. \(x = \frac{1}{2}\log_{10}\left(\frac{1}{3}y\right)\) | A1 | For correct answer for \(x\) |
| Method B: \(\frac{1}{3}y = 10^{2x}\) | M1 | For correct division of both sides by 3 |
| \(\log\frac{1}{3}y = \log 10^{2x}\) | M1 | For taking logs of both sides |
| \(\log\frac{1}{3}y = 2x\log 10\) | A1 | For correct linear equation involving logs |
| i.e. \(x = \frac{1}{2}\log_{10}\left(\frac{1}{3}y\right)\) | A1 | For correct answer for \(x\) |
| Method C: \(y = 3 \times 10^{2x} \Rightarrow \log y = \log 3 \times 10^{2x}\) | M1 | For introducing logs throughout |
| \(\log y = \log 3 + \log 10^{2x}\) | A1 | For correct RHS \(\log 3 + \log 10^{2x}\) |
| \(\log y = \log 3 + 2x\log 10\) | M1 | For correct use of \(\log a^c = b\log a\) |
| i.e. \(x = \frac{1}{2}\log_{10}\left(\frac{1}{3}y\right)\) | A1 | For correct answer for \(x\) |
| Method D: \(x = a\log(b \times 3 \times 10^{2x})\) | M1 | For substituting for \(y\), and separating RHS into at least 2 terms |
| \(x = a\log 3b + a\log 10^{2x}\) | M1 | For attempting values for \(a\) and \(b\) |
| \(x = 2ax\log 10 \Rightarrow 2a = 1 \Rightarrow a = \frac{1}{2}\) | A1 | For obtaining \(a = \frac{1}{2}\) |
| \(a\log 3b = 0 \Rightarrow 3b = 1 \Rightarrow b = \frac{1}{3}\) | A1 | For obtaining \(b = \frac{1}{3}\) |
**(i)** $\log_5\left(\frac{15 \times 20}{12}\right) = \log_5 25 = 2$ | M1, A1, A1 | For any relevant combination of $\log a \pm \log b$; For $\log 25$ – must follow correct working only; For correct answer 2
**(ii)** Method A: $\frac{1}{3}y = 10^{2x}$ | M1, M1 | For correct division of both sides by 3; For relevant use of $a = b^c \Rightarrow c = \log_a b$
Hence $2x = \log_{10}\left(\frac{1}{3}y\right)$ | A1 | For correct equation involving logs to base 10
i.e. $x = \frac{1}{2}\log_{10}\left(\frac{1}{3}y\right)$ | A1 | For correct answer for $x$
Method B: $\frac{1}{3}y = 10^{2x}$ | M1 | For correct division of both sides by 3
$\log\frac{1}{3}y = \log 10^{2x}$ | M1 | For taking logs of both sides
$\log\frac{1}{3}y = 2x\log 10$ | A1 | For correct linear equation involving logs
i.e. $x = \frac{1}{2}\log_{10}\left(\frac{1}{3}y\right)$ | A1 | For correct answer for $x$
Method C: $y = 3 \times 10^{2x} \Rightarrow \log y = \log 3 \times 10^{2x}$ | M1 | For introducing logs throughout
$\log y = \log 3 + \log 10^{2x}$ | A1 | For correct RHS $\log 3 + \log 10^{2x}$
$\log y = \log 3 + 2x\log 10$ | M1 | For correct use of $\log a^c = b\log a$
i.e. $x = \frac{1}{2}\log_{10}\left(\frac{1}{3}y\right)$ | A1 | For correct answer for $x$
Method D: $x = a\log(b \times 3 \times 10^{2x})$ | M1 | For substituting for $y$, and separating RHS into at least 2 terms
$x = a\log 3b + a\log 10^{2x}$ | M1 | For attempting values for $a$ and $b$
$x = 2ax\log 10 \Rightarrow 2a = 1 \Rightarrow a = \frac{1}{2}$ | A1 | For obtaining $a = \frac{1}{2}$
$a\log 3b = 0 \Rightarrow 3b = 1 \Rightarrow b = \frac{1}{3}$ | A1 | For obtaining $b = \frac{1}{3}$
---7 (i) Evaluate $\log _ { 5 } 15 + \log _ { 5 } 20 - \log _ { 5 } 12$.\\
(ii) Given that $y = 3 \times 10 ^ { 2 x }$, show that $x = a \log _ { 10 } ( b y )$, where the values of the constants $a$ and $b$ are to be found.
\hfill \mbox{\textit{OCR C2 2005 Q7 [7]}}