| Date | May 2016 | Marks available | 6 | Reference code | 16M.2.hl.TZ2.3 |
| Level | HL only | Paper | 2 | Time zone | TZ2 |
| Command term | Solve | Question number | 3 | Adapted from | N/A |
Question
Solve the simultaneous equations
\[\ln \frac{y}{x} = 2\]
\[\ln {x^2} + \ln {y^3} = 7.\]
Markscheme
METHOD 1
\(\ln \frac{y}{x} = 2 \Rightarrow \ln x + \ln y = 2\) A1
\(\ln {x^2} + \ln {y^3} = 7 \Rightarrow 2\ln x + 3\ln y = 7\) (M1)A1
attempting to solve for \(x\) and \(y\) \(\left( {{\text{to obtain }}\ln x = \frac{1}{5}{\text{ and }}\ln y = \frac{{11}}{5}} \right)\) (M1)
\(x = {{\text{e}}^{\frac{1}{5}}}{\text{ }}( = 1.22)\) A1
\(y = {{\text{e}}^{\frac{{11}}{5}}}{\text{ }}( = 9.03)\) A1
METHOD 2
\(\ln \frac{y}{x} = 2 \Rightarrow y = {{\text{e}}^2}x\) A1
\(\ln {x^2} + \ln {{\text{e}}^6}{x^3} = 7\) (M1)A1
attempting to solve for \(x\) (M1)
\(x = {{\text{e}}^{\frac{1}{5}}}{\text{ }}( = 1.22)\) A1
\(y = {{\text{e}}^{\frac{{11}}{5}}}{\text{ }}( = 9.03)\) A1
METHOD 3
\(\ln \frac{y}{x} = 2 \Rightarrow y = {{\text{e}}^2}x\) A1
\(\ln {x^2} + \ln {y^3} = 7 \Rightarrow \ln ({x^2}{y^3}) = 7\) A1
\({x^2}{y^3} = {{\text{e}}^7}\) (M1)
substituting \(y = {e^2}x\) into \({x^2}{y^3} = {{\text{e}}^7}\) (to obtain \({{\text{e}}^6}{x^5} = {{\text{e}}^7}\)) M1
\(x = {{\text{e}}^{\frac{1}{5}}}{\text{ }}( = 1.22)\) A1
\(y = {{\text{e}}^{\frac{{11}}{5}}}{\text{ }}( = 9.03)\) A1
[6 marks]
Examiners report
Reasonably well done. Candidates who did not obtain the correct solution generally made an error when attempting to apply logarithmic or exponential laws and hence made erroneous substitutions.
