Moderate -0.3 Part (i) requires direct application of the definition of logarithms (x^4 = 324) and simplifying a surd—straightforward with no problem-solving. Part (ii) uses the quotient law of logarithms and rearranging, which is standard bookwork. Both parts are routine exercises testing basic log laws with minimal steps, making this slightly easier than average.
11. (i) Given that \(x\) is a positive real number, solve the equation
$$\log _ { x } 324 = 4$$
writing your answer as a simplified surd.
(ii) Given that
$$\log _ { a } ( 5 y - 4 ) - \log _ { a } ( 2 y ) = 3 \quad y > 0.8,0 < a < 1$$
express \(y\) in terms of \(a\).
'Undoes' the log \(\Rightarrow x^4 = 324\); condone slips on the 324
\(\Rightarrow x = 324^{\frac{1}{4}}\)
dM1
Undoes the power correctly; \(x^4 = 324 \Rightarrow x = 324^{\frac{1}{4}}\) or alternatively \(x^4 = 324 \Rightarrow x^2 = 18 \Rightarrow x = \sqrt{18}\); may be implied by decimal answer such as awrt 4.24
First M1: correct application of subtraction law of logs (could also be awarded for addition law if student proceeds \(\log_a(a^3) + \log_a(2y)\)). Second M1: use of \(3 = \log_a a^3\); may be implied by 'undoing' the logs
A correct equation in \(a\) and \(y\); look for variations on \(\left(\frac{5y-4}{2y}\right) = a^3\)
\(\Rightarrow 5y - 4 = 2ya^3 \Rightarrow y = \dfrac{4}{5-2a^3}\)
M1A1
M1: correct attempt to change subject — cross multiplication, collecting terms in \(y\), factorising and division. A1: \(y = \dfrac{4}{5-2a^3}\) or equivalent such as \(y = \dfrac{-4}{2a^3-5}\) or \(y = \dfrac{1}{1.25-0.5a^3}\); condone fractions within fractions and isw after correct answer
# Question 11:
## Part (i):
| Working | Mark | Guidance |
|---------|------|----------|
| $\log_x 324 = 4 \Rightarrow x^4 = 324$ | M1 | 'Undoes' the log $\Rightarrow x^4 = 324$; condone slips on the 324 |
| $\Rightarrow x = 324^{\frac{1}{4}}$ | dM1 | Undoes the power correctly; $x^4 = 324 \Rightarrow x = 324^{\frac{1}{4}}$ or alternatively $x^4 = 324 \Rightarrow x^2 = 18 \Rightarrow x = \sqrt{18}$; may be implied by decimal answer such as awrt 4.24 |
| $\Rightarrow x = 3\sqrt{2}$ | A1 | Ignore any reference to $x = -3\sqrt{2}$ |
## Part (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $\log_a(5y-4) - \log_a(2y) = 3 \Rightarrow \log_a\left(\frac{5y-4}{2y}\right) = \log_a a^3$ | M1, M1 | First M1: correct application of subtraction law of logs (could also be awarded for addition law if student proceeds $\log_a(a^3) + \log_a(2y)$). Second M1: use of $3 = \log_a a^3$; may be implied by 'undoing' the logs |
| $\Rightarrow \left(\frac{5y-4}{2y}\right) = a^3$ | A1 | A correct equation in $a$ and $y$; look for variations on $\left(\frac{5y-4}{2y}\right) = a^3$ |
| $\Rightarrow 5y - 4 = 2ya^3 \Rightarrow y = \dfrac{4}{5-2a^3}$ | M1A1 | M1: correct attempt to change subject — cross multiplication, collecting terms in $y$, factorising and division. A1: $y = \dfrac{4}{5-2a^3}$ or equivalent such as $y = \dfrac{-4}{2a^3-5}$ or $y = \dfrac{1}{1.25-0.5a^3}$; condone fractions within fractions and isw after correct answer |
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11. (i) Given that $x$ is a positive real number, solve the equation
$$\log _ { x } 324 = 4$$
writing your answer as a simplified surd.\\
(ii) Given that
$$\log _ { a } ( 5 y - 4 ) - \log _ { a } ( 2 y ) = 3 \quad y > 0.8,0 < a < 1$$
express $y$ in terms of $a$.
\hfill \mbox{\textit{Edexcel C12 2019 Q11 [8]}}