Moderate -0.3 Part (a) is direct application of logarithm definition (y=2^(-3)=1/8). Part (b) requires simplifying logarithms to get 9/log₂x = log₂x, leading to a quadratic (log₂x)²=9, giving x=8 or x=1/8. Standard C2 logarithm manipulation with straightforward algebraic steps, slightly easier than average due to clean arithmetic.
\((\log_2 x)^2 = ...\) or \((\log_2 x)(\log_2 x) = ...\) (May not be seen explicitly, so M1 may be implied by later work, and the base may be 10 rather than 2)
M1
\(\log_2 x = 3 \Rightarrow x = 2^3 = 8\)
A1
\(\log_2 x = -3 \Rightarrow x = 2^{-3} = \frac{1}{8}\)
A1ft
Guidance:
Part (a):
M1 for getting out of logs correctly.
If done by change of base, \(\log_{10} y = -0.903...\) is insufficient for the M1, but \(y = 10^{-0.903}\) scores M1.
A1 for the exact answer, e.g. \(\log_{10} y = -0.903 \Rightarrow y = 0.12502...\) scores M1 (implied) A0.
Correct answer with no working scores both marks.
Allow both marks for implicit statements such as \(\log_2 0.125 = -3\).
Part (b):
\(1^{\text{st}}\) M1 for expressing 32 or 16 or 512 as a power of 2, or for a change of base enabling evaluation of \(\log_2 32\), \(\log_2 16\) or \(\log_2 512\) by calculator. (Can be implied by 5, 4 or 9 respectively).
\(1^{\text{st}}\) A1 for 9 (exact).
\(2^{\text{nd}}\) M1 for getting \((\log_2 x)^2 = \) constant. The constant can be a log or a sum of logs.
If written as \(\log_2 x^2\) instead of \((\log_2 x)^2\), allow the M mark only if subsequent work implies correct interpretation.
\(2^{\text{nd}}\) A1 for 8 (exact). Change of base methods leading to a non-exact answer score A0.
\(3^{\text{rd}}\) A1ft for an answer of \(\frac{1}{\text{their } 8}\). An ft answer may be non-exact.
Possible mistakes:
\(\log_2(2^7) = \log_2(x^2) \Rightarrow x^2 = 2^y \Rightarrow x = ...\) scores M1A1(implied by 9)M0A0
\(\log_2 512 = \log_2 x \times \log_2 x \Rightarrow x^2 = 512 \Rightarrow x = ...\) scores M0A0(9 never seen)M1A0A0
\(\log_2 48 = (\log_2 x)^2 \Rightarrow (\log_2 x)^2 = 5.585 \Rightarrow x = 5.145, x = 0.194\) scores M0A0M1A0A1ft
No working (or 'trial and improvement'):
\(x = 8\) scores M0 A0 M1 A1 A0
**Part (a)**
$\log_2 y = -3 \Rightarrow y = 2^{-3}$ | M1 |
$y = \frac{1}{8}$ or $0.125$ | A1 |
**Part (b)**
$32 = 2^5$ or $16 = 2^4$ or $512 = 2^9$ | M1 |
[or $\log_2 32 = 5 \log_2 2$ or $\log_2 16 = 4 \log_2 2$ or $\log_2 512 = 9 \log_2 2$] | |
[or $\log_2 32 = \frac{\log_{10} 32}{\log_{10} 2}$ or $\log_2 16 = \frac{\log_{10} 16}{\log_{10} 2}$ or $\log_2 512 = \frac{\log_{10} 512}{\log_{10} 2}$] | |
$\log_2 32 + \log_2 16 = 9$ | A1 |
$(\log_2 x)^2 = ...$ or $(\log_2 x)(\log_2 x) = ...$ (May not be seen explicitly, so M1 may be implied by later work, and the base may be 10 rather than 2) | M1 |
$\log_2 x = 3 \Rightarrow x = 2^3 = 8$ | A1 |
$\log_2 x = -3 \Rightarrow x = 2^{-3} = \frac{1}{8}$ | A1ft |
**Guidance:**
**Part (a):**
M1 for getting out of logs correctly.
If done by change of base, $\log_{10} y = -0.903...$ is insufficient for the M1, but $y = 10^{-0.903}$ scores M1.
A1 for the exact answer, e.g. $\log_{10} y = -0.903 \Rightarrow y = 0.12502...$ scores M1 (implied) A0.
Correct answer with no working scores both marks.
Allow both marks for implicit statements such as $\log_2 0.125 = -3$.
**Part (b):**
$1^{\text{st}}$ M1 for expressing 32 or 16 or 512 as a power of 2, or for a change of base enabling evaluation of $\log_2 32$, $\log_2 16$ or $\log_2 512$ by calculator. (Can be implied by 5, 4 or 9 respectively).
$1^{\text{st}}$ A1 for 9 (exact).
$2^{\text{nd}}$ M1 for getting $(\log_2 x)^2 = $ constant. The constant can be a log or a sum of logs.
If written as $\log_2 x^2$ instead of $(\log_2 x)^2$, allow the M mark only if subsequent work implies correct interpretation.
$2^{\text{nd}}$ A1 for 8 (exact). Change of base methods leading to a non-exact answer score A0.
$3^{\text{rd}}$ A1ft for an answer of $\frac{1}{\text{their } 8}$. An ft answer may be non-exact.
**Possible mistakes:**
$\log_2(2^7) = \log_2(x^2) \Rightarrow x^2 = 2^y \Rightarrow x = ...$ scores M1A1(implied by 9)M0A0
$\log_2 512 = \log_2 x \times \log_2 x \Rightarrow x^2 = 512 \Rightarrow x = ...$ scores M0A0(9 never seen)M1A0A0
$\log_2 48 = (\log_2 x)^2 \Rightarrow (\log_2 x)^2 = 5.585 \Rightarrow x = 5.145, x = 0.194$ scores M0A0M1A0A1ft
**No working (or 'trial and improvement'):**
$x = 8$ scores M0 A0 M1 A1 A0
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