C3 June 2010 Q8
8.
Given that
\[\ln(2x^2 + 9x - 5) = 1 + \ln(x^2 + 2x - 15), \quad x \ne -5,\]| Scheme | Marks |
|---|---|
| \(\dfrac{(x + 5)(2x - 1)}{(x + 5)(x - 3)} = \dfrac{(2x - 1)}{(x - 3)}\) | M1 B1 A1 aef |
| (3) |
Notes
(a) M1: An attempt to factorise the numerator.
B1: Correct factorisation of denominator to give \((x + 5)(x - 3)\). Can be seen anywhere.
| Scheme | Marks |
|---|---|
| \(\ln\left(\dfrac{2x^2 + 9x - 5}{x^2 + 2x - 15}\right) = 1\) | M1 |
| \(\dfrac{2x^2 + 9x - 5}{x^2 + 2x - 15} = \mathrm{e}\) | dM1 |
| \(\dfrac{2x - 1}{x - 3} = \mathrm{e} \Rightarrow 3\mathrm{e} - 1 = x(\mathrm{e} - 2)\) | M1 |
| \(\Rightarrow x = \dfrac{3\mathrm{e} - 1}{\mathrm{e} - 2}\) | A1 aef cso |
| (4) | |
| (7 marks) |
Notes
(b) M1: Uses a correct law of logarithms to combine at least two terms. This usually is achieved by the subtraction law of logarithms to give \(\ln\left(\dfrac{2x^2 + 9x - 5}{x^2 + 2x - 15}\right) = 1\).
The product law of logarithms can be used to achieve \(\ln(2x^2 + 9x - 5) = \ln\left(\mathrm{e}(x^2 + 2x - 15)\right)\).
The product and quotient law could also be used to achieve \(\ln\left(\dfrac{2x^2 + 9x - 5}{\mathrm{e}(x^2 + 2x - 15)}\right) = 0\).
dM1: Removing ln’s correctly by the realisation that the anti-ln of 1 is e. Note that this mark is dependent on the previous method mark being awarded.
M1: Collect \(x\) terms together and factorise. Note that this is not a dependent method mark.
A1: \(\dfrac{3\mathrm{e} - 1}{\mathrm{e} - 2}\) or \(\dfrac{3\mathrm{e}^1 - 1}{\mathrm{e}^1 - 2}\) or \(\dfrac{1 - 3\mathrm{e}}{2 - \mathrm{e}}\). aef
Note that the answer needs to be in terms of e. The decimal answer is 9.9610559… Note that the solution must be correct in order for you to award this final accuracy mark.
Note: See Appendix for an alternative method of long division.