Solve the equation ${4^{x - 1}} = {2^x} + 8$.

${2^{2x - 2}} = {2^x} + 8$     (M1) 

$\frac{1}{4}{2^{2x}} = {2^x} + 8$     (A1)

${2^{2x}} - 4 \times {2^x} - 32 = 0$     A1

$({2^x} - 8)({2^x} + 4) = 0$     (M1)

${2^x} = 8 \Rightarrow x = 3$     A1

Notes: Do not award final A1 if more than 1 solution is given.

[5 marks]

Very few candidates knew how to solve this equation. A significant number guessed the answer using trial and error after failed attempts to solve it. A number of misconceptions were identified involving properties of logarithms and exponentials.