| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Combine logs into single logarithm |
| Difficulty | Moderate -0.8 This is a straightforward application of logarithm laws (power rule and addition/subtraction rules) followed by simple substitution. Part (i) requires routine manipulation of logs, and part (ii) is direct verification by substituting x=4 into a simplified expression. Both parts are mechanical with no problem-solving or insight required, making this easier than average. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\log_{10}\frac{(x+4)(x+16)}{x^2}\) | B1 B1 | 2 marks |
| (ii) Substitute \(x = 4\) gives \(\log_{10}\frac{(x+4)(x+16)}{x^2} = \log_{10}\frac{8 \times 20}{16} = \log_{10}10 = 1\) | M1 E1 | 2 marks |
**(i)** $\log_{10}\frac{(x+4)(x+16)}{x^2}$ | B1 B1 | 2 marks | Numerator Denominator
**(ii)** Substitute $x = 4$ gives $\log_{10}\frac{(x+4)(x+16)}{x^2} = \log_{10}\frac{8 \times 20}{16} = \log_{10}10 = 1$ | M1 E1 | 2 marks3 (i) Write $\log _ { 10 } ( x + 4 ) - 2 \log _ { 10 } x + \log _ { 10 } ( x + 16 )$ as a single logarithm.\\
(ii) Without using your calculator, verify that $x = 4$ is a root of the equation
$$\log _ { 10 } ( x + 4 ) - 2 \log _ { 10 } x + \log _ { 10 } ( x + 16 ) = 1$$
\hfill \mbox{\textit{OCR MEI C2 Q3 [4]}}