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 \(2 =  - x\).