C3 June 2013 Q6
6. Find algebraically the exact solutions to the equations
Give your answer to (b) in the form \(\dfrac{a+\ln b}{c+\ln d}\) where \(a\), \(b\), \(c\) and \(d\) are integers. (5)
| Scheme | Marks |
|---|---|
| \(\ln(4-2x)(9-3x),=\ln(x+1)^2\) | M1, M1 |
| So \(36-30x+6x^2=x^2+2x+1\) and \(5x^2-32x+35=0\) | A1 |
| Solve \(5x^2-32x+35=0\) to give \(x=\dfrac{7}{5}\) oe ( Ignore the solution \(x=5\)) | M1A1 |
| (5) |
Notes
M1 Uses addition law on lhs of equation. Accept slips on the signs. If one of the terms is taken over to the rhs it would be for the subtraction law.
M1 Uses power rule for logs write the \(2\ln(x+1)\) term as \(\ln(x+1)^2\). Condone invisible brackets
A1 Undoes the logs to obtain the 3TQ =0. \(5x^2-32x+35=0\). Accept equivalences. The equals zero may be implied by a subsequent solution of the equation.
M1 Solves a quadratic by any allowable method.
The quadratic cannot be a version of \((4-2x)(9-3x)=0\) however.
A1 Deduces \(x=1.4\) or equivalent. Accept both \(x=1.4\) and \(x=5\). Candidates do not have to eliminate \(x=5\).
You may ignore any other solution as long as it is not in the range \(-1<x<2\). Extra solutions in the range scores A0.
| Scheme | Marks |
|---|---|
| Take \(\log_{\mathrm{e}}\)’s to give \(\ln 2^x+\ln\mathrm{e}^{3x+1}=\ln 10\) | M1 |
| \(x\ln 2+(3x+1)\ln\mathrm{e}=\ln 10\) | M1 |
| \(x(\ln 2+3\ln\mathrm{e})=\ln 10-\ln\mathrm{e}\Rightarrow x=..\) | dM1 |
| and uses \(\ln\mathrm{e}=1\) | M1 |
| \(x=\dfrac{-1+\ln 10}{3+\ln 2}\) | A1 |
| Note that the 4th M mark may occur on line 2 | |
| (5) | |
| (10 marks) |
Notes
M1 Takes logs of both sides and splits LHS using addition law. If one of the terms is taken to the other side it can be awarded for taking logs of both sides and using the subtraction law.
M1 Taking both powers down using power rule. It is not wholly dependent upon the first M1 but logs of both sides must have been taken. Below is an example of M0M1
\(\ln 2^x\times\ln\mathrm{e}^{3x+1}=\ln 10\Rightarrow x\ln 2\times(3x+1)\ln e=\ln 10\)
dM1 This is dependent upon both previous two M’s being scored. It can be awarded for a full method to solve their linear equation in \(x\). The terms in \(x\) must be collected on one side of the equation and factorised. You may condone slips in signs for this mark but the process must be correct and leading to \(x=\ldots\)
M1 Uses \(\ln\mathrm{e}=1\). This could appear in line 2, but it must be part of their equation and not just a statement.
Another example where it could be awarded is \(\mathrm{e}^{3x+1}=\dfrac{10}{2^x}\Rightarrow 3x+1=\ldots\)
A1 Obtains answer \(x=\dfrac{-1+\ln 10}{3+\ln 2}=\left(\dfrac{\ln 10-1}{3+\ln 2}\right)=\left(\dfrac{\log_{\mathrm{e}}10-1}{3+\log_{\mathrm{e}}2}\right)oe\). DO NOT ISW HERE
Note 1: If the candidate takes \(\log_{10}\)’s of both sides can score M1M1dM1M0A0 for 3 out of 5.
Answer = \(x=\dfrac{-\log\mathrm{e}+\log 10}{3\log\mathrm{e}+\log 2}=\left(\dfrac{-\log\mathrm{e}+1}{3\log\mathrm{e}+\log 2}\right)\)
Note 2: If the candidate writes \(x=\dfrac{-1+\log 10}{3+\log 2}\) without reference to natural logs then award M4 but with hold the last A1 mark, scoring 4 out of 5.
Alt 1 to 6(b)
| Scheme | Marks |
|---|---|
| Writes lhs in e’s \(\ 2^x\mathrm{e}^{3x+1}=10\Rightarrow\mathrm{e}^{x\ln 2}\mathrm{e}^{3x+1}=10\) | 1st M1 |
| \(\Rightarrow\mathrm{e}^{x\ln 2+3x+1}=10,\quad x\ln 2+3x+1=\ln 10\) | 2nd M1, 4th M1 |
| \(x(\ln 2+3)=\ln 10-1\Rightarrow x=..\) | dM1 |
| \(x=\dfrac{-1+\ln 10}{3+\ln 2}\) | A1 |
| (5) |
M1 Writes the lhs of the expression in e’s. Seeing \(2^x=\mathrm{e}^{x\ln 2}\) in their equation is sufficient
M1 Uses the addition law on the lhs to produce a single exponential
dM1 Takes ln’s of both sides to produce and attempt to solve a linear equation in \(x\)
You may condone slips in signs for this mark but the process must be correct leading to \(x=..\)
M1 Uses \(\ln\mathrm{e}=1\). This could appear in line 2