Solve the equation ${4^x} + {2^{x + 2}} = 3$.

attempt to form a quadratic in ${2^x}$     M1

${({2^x})^2} + 4 \bullet {2^x} - 3 = 0$    A1

${2^x} = \frac{{ - 4 \pm \sqrt {16 + 12} }}{2}{\text{ }}\left( { =  - 2 \pm \sqrt 7 } \right)$    M1

${2^x} =  - 2 + \sqrt 7 {\text{ }}\left( {{\text{as }} - 2 - \sqrt 7  < 0} \right)$    R1

$x = {\log _2}\left( { - 2 + \sqrt 7 } \right){\text{ }}\left( {x = \frac{{\ln \left( { - 2 + \sqrt 7 } \right)}}{{\ln 2}}} \right)$    A1

Note: Award R0 A1 if final answer is $x = {\log _2}\left( { - 2 + \sqrt 7 } \right)$.

[5 marks]