Easy -1.8 This is a straightforward index law question requiring only recognition that 32 = 2^5 and √2 = 2^(1/2), then adding exponents. It's simpler than the calibration example at -1.5 as it's single-step with no multi-part structure, making it below-average difficulty even for C1.
Must be explicitly seen: \(32\sqrt{2}=2^a\). Even writing \(32\times2=2^5\times2(=2^6)\) is OK but simply writing \(32\times2=2^6\) is NOT
\(\sqrt{2}=2^{\frac{1}{2}}\) or \(\sqrt{2048}=(2048)^{\frac{1}{2}}\)
B1
May be implied. Special case: if \(\sqrt{2}=2^{\frac{1}{2}}\) not explicitly seen but final answer includes \(\frac{1}{2}\), second B1 given by implication
\(a = \dfrac{11}{2}\) (or \(5\dfrac{1}{2}\) or \(5.5\))
B1
Need \(a=\ldots\) so \(2^{\frac{11}{2}}\) is B0. \(a=5.5\) seen awards 3/3 unless clearly from incorrect work
# Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $32 = 2^5$ or $2048 = 2^{11}$ | B1 | Must be explicitly seen: $32\sqrt{2}=2^a$. Even writing $32\times2=2^5\times2(=2^6)$ is OK but simply writing $32\times2=2^6$ is NOT |
| $\sqrt{2}=2^{\frac{1}{2}}$ or $\sqrt{2048}=(2048)^{\frac{1}{2}}$ | B1 | May be implied. Special case: if $\sqrt{2}=2^{\frac{1}{2}}$ not explicitly seen but final answer includes $\frac{1}{2}$, second B1 given by implication |
| $a = \dfrac{11}{2}$ (or $5\dfrac{1}{2}$ or $5.5$) | B1 | Need $a=\ldots$ so $2^{\frac{11}{2}}$ is B0. $a=5.5$ seen awards 3/3 unless clearly from incorrect work |
---