| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Session | Specimen |
| Marks | 4 |
| Topic | Laws of Logarithms |
| Type | Evaluate log expression using laws |
| Difficulty | Easy -1.2 This is a straightforward application of basic logarithm laws (ln(ab) = ln(a) + ln(b) and ln(a/b) = ln(a) - ln(b)) with given values. It requires only direct substitution and arithmetic: ln(36) = ln(12×3) = ln(12) + ln(3), and ln(0.5) = ln(6/12) = ln(6) - ln(12) where ln(6) = ln(3×2) can be found from the given values. This is simpler than typical A-level questions as it involves no algebraic manipulation or problem-solving—just mechanical application of laws with provided numerical values. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules |
$\ln(36) = \ln(12 \times 3) = \ln(12) + \ln(3)$ M1
$= 2.484907 + 1.098612$
$= 3.583519$
$= 3.5835$ A1
$\ln(0.5) = -\ln(2) = -\ln(\sqrt{4})$ M1
$= -\frac{1}{2}\ln(4) = -\frac{1}{2}(\ln(12) - \ln(3))$
$= -(2.484907 - 1.098612)/2$
$= -0.6931$ [There are several alternative routes to the final answers.] A1
**Total: 4 marks**2 You are given that $\ln ( 12 ) = 2.484907$ and $\ln ( 3 ) = 1.098612$, correct to 6 decimal places. Use the laws of logarithms to obtain the values of $\ln ( 36 )$ and $\ln ( 0.5 )$, correct to 4 decimal places. You must show your numerical working.
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 Q2 [4]}}