4MA0 Higher June 2016 Paper 3R Q21
21 Solve \(3 \times 4^{2k + 8} = 24\)
Show your working clearly.
(4)
| Scheme | Marks |
|---|---|
| \((2^2)^{2k + 8} = 2^3\) or \(4^{\frac{3}{2}} = 8\) or \(2^{4k + 16} = 2^3\) or \(4^{\frac{3}{2}} = 4^{2k + 8}\) | M2 |
| \(4k + 16 = 3\) or \(2k + 8 = 1.5\) oe | M1 |
| \(-\dfrac{13}{4}\) | A1oe |
| (4) |
Notes
M2: M1 for \(4^{2k + 8} = 8\) or \(3 \times 4^{\frac{3}{2}} = 24\)
M1: A correct equation in \(k\) or a fully correct method to find \(k\)
A1oe: Dep on at least M2