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]