Write the expression $3\ln 2 - \ln 4$ in the form $\ln k$, where $k \in \mathbb{Z}$.

Hence or otherwise, solve $3\ln 2 - \ln 4 =  - \ln x$.

correct application of $\ln {a^b} = b\ln a$ (seen anywhere)     (A1)

eg$\;\;\;\ln 4 = 2\ln 2,{\text{ }}3\ln 2 = \ln {2^3},{\text{ }}3\log 2 = \log 8$

correct working     (A1)

eg$\;\;\;3\ln 2 - 2\ln 2,{\text{ }}\ln 8 - \ln 4$

$\ln 2\;\;\;{\text{(accept }}k = 2{\text{)}}$     A1     N2

[3 marks]

METHOD 1

attempt to substitute their answer into the equation     (M1)

eg$\;\;\;\ln 2 =  - \ln x$

correct application of a log rule     (A1)

eg$\;\;\;\ln \frac{1}{x},{\text{ }}\ln \frac{1}{2} = \ln x,{\text{ }}\ln 2 + \ln x = \ln 2x\;\;\;( = 0)$

$x = \frac{1}{2}$     A1     N2

METHOD 2

attempt to rearrange equation, with  $3\ln 2$ written as $\ln {2^3}$ or $\ln 8$     (M1)

eg$\;\;\;\ln x = \ln 4 - \ln {2^3},{\text{ }}\ln 8 + \ln x = \ln 4,{\text{ }}\ln {2^3} = \ln 4 - \ln x$

correct working applying $\ln a \pm \ln b$     (A1)

eg$\;\;\;\frac{4}{8},{\text{ }}8x = 4,{\text{ }}\ln {2^3} = \ln \frac{4}{x}$

$x = \frac{1}{2}$     A1     N2

[3 marks]

Total [6 marks]

Part (a) was answered correctly by a large number of candidates, though there were quite a few who applied the rules of logarithms in the wrong order.

In part (b), many candidates knew to set their answer from part (a) equal to $- \ln x$, but then a good number incorrectly said that $\ln 2 =  - \ln x$ led to $2 =  - x$.