Higher November 2022 Paper 6 Q9
9 Given that \((2^k)^6 \times 8 = 2^{45}\), find the value of \(k\). [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 7 nfww | 3 | M2 for 6\(k\) + 3 [= 45] or M1 for 6\(k\) or \(2^3\) soi Alternative method: M2 for \(2^k = \sqrt[6]{2^{42}}\) or \(2^k\) = 128 or M1 for \((2^k)^6\) = any of the below \(\frac{2^{45}}{2^3}\), \(2^{42}\), \(4.398\ldots \times 10^{12}\), \(4.4[0] \times 10^{12}\) | Could be implied by manipulation of powers followed by ÷ 6 Accept \(\frac{2^{45}}{2^3}\), \(4.398\ldots \times 10^{12}\) or \(4.4[0] \times 10^{12}\) in place of \(2^{42}\) Do not accept \((2^k)^6 = \frac{2^{45}}{8}\) |