Standard +0.3 This is a straightforward logarithm equation requiring application of standard log laws (power law and subtraction law) to simplify, then solving the resulting quadratic. It's slightly above average difficulty due to the algebraic manipulation needed after applying log laws, but remains a standard textbook exercise with no novel insight required.
Use law for the logarithm of a product, quotient or power
M1
Condone \(\ln\dfrac{x}{x-1}\) for M1
Obtain a correct equation free of logarithms
A1
e.g. \((2x-3)(x-1) = x^2\) or \(x^2 - 5x + 3 = 0\)
Solve a 3-term quadratic obtaining at least one root
M1
Must see working if using an incorrect quadratic \(\left(\dfrac{5 \pm \sqrt{13}}{2}\right)\)
Obtain answer \(x = 4.30\) only
A1
Q asks for 2 d.p. Do not ISW. Overspecified answers score A0. Overspecified and no working can score M1A0
Total
4
**Question 2:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use law for the logarithm of a product, quotient or power | M1 | Condone $\ln\dfrac{x}{x-1}$ for M1 |
| Obtain a correct equation free of logarithms | A1 | e.g. $(2x-3)(x-1) = x^2$ or $x^2 - 5x + 3 = 0$ |
| Solve a 3-term quadratic obtaining at least one root | M1 | Must see working if using an incorrect quadratic $\left(\dfrac{5 \pm \sqrt{13}}{2}\right)$ |
| Obtain answer $x = 4.30$ only | A1 | Q asks for 2 d.p. Do not ISW. Overspecified answers score A0. Overspecified and no working can score M1A0 |
| **Total** | **4** | |
2 Showing all necessary working, solve the equation $\ln ( 2 x - 3 ) = 2 \ln x - \ln ( x - 1 )$. Give your answer correct to 2 decimal places.\\
\hfill \mbox{\textit{CAIE P3 2019 Q2 [4]}}