16N.1.AHL.TZ0.H_7

pestleMathematicsAIHLPaper 116N· sl-1-5-intro-to-logssource ↗

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

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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]

Examiners’ report

[N/A]