| Date | May 2010 | Marks available | 5 | Reference code | 10M.1.hl.TZ1.4 |
| Level | HL only | Paper | 1 | Time zone | TZ1 |
| Command term | Solve | Question number | 4 | Adapted from | N/A |
Question
Solve the equation \({4^{x - 1}} = {2^x} + 8\).
Markscheme
\({2^{2x - 2}} = {2^x} + 8\) (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]
Examiners report
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.
