Higher June 2025 Paper 3 Q11
11 \(\dfrac{3^n \times 3^{20}}{3^8} = 3^{14}\)
(a) Find the value of \(n\). (2)
(b) Write \(\left(64m^{12}\right)^{\frac{2}{3}}\) in the form \(am^b\) where \(a\) and \(b\) are integers. (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 2 | M1 | for a correct start, using one rule of indices, eg \(3^n \times 3^{12} = 3^{14}\) or \(3^n \times 3^{20} = 3^{22}\) or \(3^{20} \div 3^8 = 3^{12}\) or \(3^{14} \times 3^8 = 3^{22}\) or \(\dfrac{3^{n + 20}}{3^8} = 3^{14}\) or \(3^{n - 8} \times 3^{20} = 3^{14}\) or for forming an equation in \(n\), eg \(n + 20 - 8 = 14\) oe or (\(n =\)) \(14 + 8 - 20\) |
| A1 | cao SCB1 for an answer of \(3^2\) if M0 scored |
Additional guidance
Allow an answer of \(3^n = 3^2\)
An answer of 9 or \(3^n = 9\), \(n \neq 2\) on its own is to be awarded 0 marks
| Answer | Mark | Mark scheme |
|---|---|---|
| \(16m^8\) | M1 | for an intention to find the cube root and square, eg \(\sqrt[3]{64m^{12}}^{\,2}\) or \(\sqrt[3]{\left(64m^{12}\right)^2}\) or \(\left(4m^4\right)^2\) or \(\sqrt[3]{4096m^{24}}\) or for \(am^8\) with \(a \neq 16\) or \(16m^b\) with \(b \neq 8\) |
| A1 | cao |
Additional guidance
Do not condone missing brackets
16 or \(m^8\) imply M1
Allow multiplication sign for M1
Accept \(a = 16\), \(b = 8\)