C3 June 2014 Q2

Notes

M1 Proceeds from \(2\ln(2x+1)-10=0\) to \(\ln(2x+1)=5\) before taking exp's to achieve \(x\) in terms of \(\mathrm{e}^5\)
Accept for M1 \(2\ln(2x+1)-10=0\Rightarrow\ln(2x+1)=5\Rightarrow x=\mathrm{f}(\mathrm{e}^5)\)
Alternatively they could use the power law before taking exp's to achieve \(x\) in terms of \(\sqrt{\mathrm{e}^{10}}\)
\(2\ln(2x+1)=10\Rightarrow\ln(2x+1)^2=10\Rightarrow(2x+1)^2=\mathrm{e}^{10}\Rightarrow x=\mathrm{g}\left(\sqrt{\mathrm{e}^{10}}\right)\)

A1 cso. Accept \(x=\dfrac{\mathrm{e}^5-1}{2}\) or other exact simplified alternatives such as \(x=\dfrac{\mathrm{e}^5}{2}-\dfrac{1}{2}\). Remember to isw.
The decimal answer of 73.7 will score M1A0 unless the exact answer has also been given.
The answer \(\dfrac{\sqrt{\mathrm{e}^{10}}-1}{2}\) does not score this mark unless simplified. \(x=\dfrac{\pm\mathrm{e}^5-1}{2}\) is M1A0

Notes

M1 Takes ln’s or logs of both sides and applies the addition law.
\(\ln(3^x\mathrm{e}^{4x})=\ln 3^x+\ln\mathrm{e}^{4x}\) or \(\ln(3^x\mathrm{e}^{4x})=\ln 3^x+4x\) is evidence for the addition law
If the \(\mathrm{e}^{4x}\) was ‘moved’ over to the right hand side score for either \(\mathrm{e}^{7-4x}\) or the subtraction law.
\(\ln\dfrac{\mathrm{e}^7}{\mathrm{e}^{4x}}=\ln\mathrm{e}^7-\ln\mathrm{e}^{4x}\) or \(3^x\mathrm{e}^{4x}=\mathrm{e}^7\Rightarrow 3^x=\dfrac{\mathrm{e}^7}{\mathrm{e}^{4x}}\Rightarrow 3^x=\mathrm{e}^{7-4x}\) is evidence of the subtraction law

M1 Uses the power law of logs (seen at least once in a term with x as the index Eg \(3^x,\mathrm{e}^{4x}\) or \(\mathrm{e}^{7-4x}\)).
\(\ln 3^x+\ln\mathrm{e}^{4x}=\ln\mathrm{e}^7\Rightarrow x\ln 3+4x\ln\mathrm{e}=7\ln\mathrm{e}\) is an example after the addition law
\(3^x=\mathrm{e}^{7-4x}\Rightarrow x\log 3=(7-4x)\log\mathrm{e}\) is an example after the subtraction law.
It is possible to score M0M1 by applying the power law after an incorrect addition/subtraction law
For example \(3^x\mathrm{e}^{4x}=\mathrm{e}^7\Rightarrow\ln(3^x)\times\ln(\mathrm{e}^{4x})=\ln\mathrm{e}^7\Rightarrow x\ln 3\times 4x\ln\mathrm{e}=7\ln\mathrm{e}\)

dM1 This is dependent upon both previous M’s. Collects/factorises out term in \(x\) and proceeds to \(x=\).
Condone sign slips for this mark. An unsimplified answer can score this mark.

A1 If the candidate has taken ln’s then they must use \(\ln\mathrm{e}=1\) and achieve \(x=\dfrac{7}{(\ln 3+4)}\) or equivalent.
If the candidate has taken log’s they must be writing log as oppose to ln and achieve \(x=\dfrac{7\log\mathrm{e}}{(\log 3+4\log\mathrm{e})}\) or other exact equivalents such as \(x=\dfrac{7\log\mathrm{e}}{\log 3\mathrm{e}^4}\).

Alt 1 2(b)

Alt 2 2(b) Using logs

Alt 3 2(b) Using \(\log_3\)

Alt 4 2(b) Using \(3^x=\mathrm{e}^{x\ln 3}\)