Solve the simultaneous equations

$$\ln \frac{y}{x} = 2$$

$$\ln {x^2} + \ln {y^3} = 7.$$

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]

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.