Edexcel AEA 2008 June — Question 5 14 marks

Exam BoardEdexcel
ModuleAEA (Advanced Extension Award)
Year2008
SessionJune
Marks14
PaperDownload PDF ↗
TopicLaws of Logarithms
TypeTwo unrelated log parts: one non-log algebraic part
DifficultyChallenging +1.8 Part (i) requires solving a system where incorrect logarithm rules accidentally hold, demanding algebraic manipulation and insight to find p=9, q=27. Part (ii) involves simplifying logarithm expressions, change of base, and solving a cubic equation. Both parts require non-routine problem-solving beyond standard A-level techniques, typical of AEA's challenging nature.
Spec1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b

  1. Anna, who is confused about the rules for logarithms, states that $$(\log_3 p)^2 = \log_3 (p^2)$$ and $$\log_3(p + q) = \log_3 p + \log_3 q.$$ However, there is a value for \(p\) and a value for \(q\) for which both statements are correct. Find the value of \(p\) and the value of \(q\). [7]
  2. Solve $$\frac{\log_3(3x^3 - 23x^2 + 40x)}{\log_3 9} = 0.5 + \log_3(3x - 8).$$ [7]

\begin{enumerate}[label=(\roman*)]
\item Anna, who is confused about the rules for logarithms, states that
$$(\log_3 p)^2 = \log_3 (p^2)$$
and
$$\log_3(p + q) = \log_3 p + \log_3 q.$$

However, there is a value for $p$ and a value for $q$ for which both statements are correct.

Find the value of $p$ and the value of $q$. [7]

\item Solve
$$\frac{\log_3(3x^3 - 23x^2 + 40x)}{\log_3 9} = 0.5 + \log_3(3x - 8).$$ [7]
\end{enumerate}

\hfill \mbox{\textit{Edexcel AEA 2008 Q5 [14]}}