Higher June 2022 Paper 2 Q24
24
\[\dfrac{18 \times \left(\sqrt{27}\right)^{4n + 6}}{6 \times 9^{2n + 8}} = 3^x\]Express \(x\) in terms of \(n\)
Show your working clearly and simplify your expression.
(3)
| Scheme | Marks |
|---|---|
eg \(\dfrac{2 \times 3 \times 3 \times \left(3^{\frac{3}{2}}\right)^{4n + 6}}{2 \times 3 \times 3^{2(2n + 8)}}\) or \(\dfrac{3 \times 3^{\frac{3}{2}(4n + 6)}}{3^{2(2n + 8)}}\) \(\sqrt{27}\) to be changed to a power of 3 and not \(3\sqrt{3}\) unless recovered | M1 |
eg \(\dfrac{3 \times 3^{6n + 9}}{3^{4n + 16}}\) or \(\dfrac{3^{6n + 10}}{3^{4n + 16}}\) or \(\dfrac{3 \times 3^{1.5(4n + 6)}}{3^{2(2n + 8)}}\) or \(\dfrac{3^2 \times 3^{6n + 9}}{3 \times 3^{4n + 16}}\) or \(\dfrac{3^{6n + 11}}{3^{4n + 17}}\) oe or eg \(3^{6n + 11} = 3^x \times 3^{4n + 17}\) oe | M1 |
| Working required Answer: \(2n - 6\) | A1 |
| (3) | |
| (3 marks) |
Notes
M1: For 2 of:
• writing 18 as \(2 \times 3^2\) oe and 6 as \(2 \times 3\) OR cancelling 6 & 18 fully
• writing \(\sqrt{27}\) as \(3^{\frac{3}{2}}\) or \(3 \times 3^{\frac{1}{2}}\) OR \(\left(\sqrt{27}\right)^{4n + 6}\) as \(\left(3^3\right)^{2n + 3}\) or \(3^{6n + 9}\)
• writing 9 as \(3^2\) OR \(9^{2n + 8}\) as \(3^{2(2n + 8)}\) or \(3^{4n + 16}\)
M1: For a correct expression or equation using only powers of 3 (powers of 3 but not necessarily a single power)
A1: oe eg \(2(n - 3)\) dep on M1