Foundation June 2022 Paper 2R Q24
24
\(\dfrac{2^k}{4^n} = 2^x\)
Find an expression for \(x\) in terms of \(k\) and \(n\)
(2)
| Scheme | Marks |
|---|---|
\((4^n =)(2^2)^n\) or \((4^n =) 2^{2n}\) oe eg \(2^k \div 2^{2n} = 2^x\) or \(2^k = 4^{\frac{1}{2}k}\) and \(2^x = 4^{\frac{1}{2}x}\) oe eg \(\dfrac{4^{\frac{1}{2}k}}{4^n} = 4^{\frac{1}{2}x}\) | M1 |
| \(k - 2n\) | A1 |
| (2) | |
| (2 marks) |
Notes
M1: for writing \(4^n\) as \((2^2)^n\) or \(2^{2n}\) or for writing each term in terms of 4 ie \(2^k = 4^{\frac{1}{2}k}\) and \(2^x = 4^{\frac{1}{2}x}\)
If these things are seen in working, award this mark even if followed by incorrect working – if not a choice of methods
A1: allow \(2^{k-2n}\)