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Date November 2016 Marks available 5 Reference code 16N.1.hl.TZ0.7
Level HL only Paper 1 Time zone TZ0
Command term Solve Question number 7 Adapted from N/A

Question

Solve the equation \({4^x} + {2^{x + 2}} = 3\).

Markscheme

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]

Syllabus sections

Topic 1 - Core: Algebra » 1.2 » Exponents and logarithms.

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