C3 June 2017 Q2
2. Find the exact solutions, in their simplest form, to the equations
| Scheme | Marks |
|---|---|
| \(\mathrm{e}^{3x - 9} = 8 \Rightarrow 3x - 9 = \ln 8\) | M1 |
| \(\Rightarrow x = \dfrac{\ln 8 + 9}{3}, = \ln 2 + 3\) | A1, A1 |
| (3) |
Notes
M1: Takes ln's of both sides and uses the power law. You may even accept candidates taking logs of both sides
A1: A correct unsimplified answer \(\dfrac{\ln 8 + 9}{3}\) or equivalent such as \(\dfrac{\ln 8\mathrm{e}^9}{3}, 3 + \ln\left(\sqrt[3]{8}\right), \dfrac{\log 8}{3\log\mathrm{e}} + 3\) or even 3.69
A1: cso \(\ln 2 + 3\). Accept \(\ln 2\mathrm{e}^3\)
Alt I (a)
\(\mathrm{e}^{3x - 9} = 8 \Rightarrow \dfrac{\mathrm{e}^{3x}}{\mathrm{e}^9} = 8 \Rightarrow \mathrm{e}^{3x} = 8\mathrm{e}^9 \Rightarrow 3x = \ln\left(8\mathrm{e}^9\right)\) for M1 (Condone slips on index work and lack of bracket)
Alt II (a)
\(\mathrm{e}^{x - 3} = \sqrt[3]{8} \Rightarrow x - 3 = \ln\left(\sqrt[3]{8}\right)\) for M1 (Condone slips on the 9. Eg \(\mathrm{e}^{x - 9} = 2 \Rightarrow x - 9 = \ln 2\))
| Scheme | Marks |
|---|---|
| \(\ln(2y + 5) = 2 + \ln(4 - y)\) | |
| \(\ln\left(\dfrac{2y + 5}{4 - y}\right) = 2\) | M1 |
| \(\left(\dfrac{2y + 5}{4 - y}\right) = \mathrm{e}^2\) | M1 |
| \(2y + 5 = \mathrm{e}^2(4 - y) \Rightarrow 2y + \mathrm{e}^2 y = 4\mathrm{e}^2 - 5 \Rightarrow y = \dfrac{4\mathrm{e}^2 - 5}{2 + \mathrm{e}^2}\) | dM1, A1 |
| (4) | |
| (7 marks) |
Notes
M1: Uses a correct method to combine two terms to create a single ln term.
Eg. Score for \(2 + \ln(4 - y) = \ln\left(\mathrm{e}^2(4 - y)\right)\) or \(\ln(2y + 5) - \ln(4 - y) = \ln\left(\dfrac{2y + 5}{4 - y}\right)\)
Condone slips on the signs and coefficients of the terms, but not on the \(\mathrm{e}^2\)
M1: Scored for an attempt to undo the ln's to get an equation in \(y\) This must be awarded after an attempt to combine the ln terms. Award for \(\ln(\mathrm{g}(y)) = 2 \Rightarrow \mathrm{g}(y) = \mathrm{e}^2\) and can be scored eg where \(\mathrm{g}(y) = 2y + 5 - (4 - y)\)
It cannot be awarded for just \(2y + 5 = \mathrm{e}^2 + 4 - y\) where the candidate attempts to undo term by term
dM1: Dependent upon both previous M's. It is for making \(y\) the subject. Expect to see both terms in \(y\) collected and factorised (may be implied) before reaching \(y\) =. Condone slips, for eg, on signs. \(y\) =2.615 scores this.
A1: \(y = \dfrac{4\mathrm{e}^2 - 5}{2 + \mathrm{e}^2}\) or equivalent such as \(y = 4 - \dfrac{13}{2 + \mathrm{e}^2}\) ISW after you see the correct answer.
Special Case: \(\ln(2y + 5) - \ln(4 - y) = 2 \Rightarrow \dfrac{\ln(2y + 5)}{\ln(4 - y)} = 2 \Rightarrow \dfrac{2y + 5}{4 - y} = \mathrm{e}^2 \Rightarrow\) Correct answer score M0 M1 M1 A0