Solve the equation \({8^{x - 1}} = {6^{3x}}\). Express your answer in terms of \(\ln 2\) and \(\ln 3\).

METHOD 1

\({2^{3(x - 1)}} = {(2 \times 3)^{3x}}\)     M1

Note:     Award M1 for writing in terms of 2 and 3.

\({2^{3x}} \times {2^{ - 3}} = {2^{3x}} \times {3^{3x}}\)

\({2^{ - 3}} = {3^{3x}}\)     A1

\(\ln \left( {{2^{ - 3}}} \right) = \ln \left( {{3^{3x}}} \right)\)     (M1)

\( - 3\ln 2 = 3x\ln 3\)     A1

\(x =  - \frac{{\ln 2}}{{\ln 3}}\)     A1

METHOD 2

\(\ln {8^{x - 1}} = \ln {6^{3x}}\)     (M1)

\((x - 1)\ln {2^3} = 3x\ln (2 \times 3)\)     M1A1

\(3x\ln 2 - 3\ln 2 = 3x\ln 2 + 3x\ln 3\)     A1

\(x =  - \frac{{\ln 2}}{{\ln 3}}\)     A1

METHOD 3

\(\ln {8^{x - 1}} = \ln {6^{3x}}\)     (M1)

\((x - 1)\ln 8 = 3x\ln 6\)     A1

\(x = \frac{{\ln 8}}{{\ln 8 - 3\ln 6}}\)     A1

\(x = \frac{{3\ln 2}}{{\ln \left( {\frac{{{2^3}}}{{{6^3}}}} \right)}}\)     M1

\(x =  - \frac{{\ln 2}}{{\ln 3}}\)     A1

[5 marks]