| Exam Board | Edexcel |
|---|---|
| Module | AEA (Advanced Extension Award) |
| Year | 2006 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Change of base or reciprocal relationship |
| Difficulty | Challenging +1.2 Part (a) is a standard proof of the reciprocal relationship for logarithms that appears in textbooks. Part (b) requires simple algebraic manipulation of the result from (a). Part (c) is more challenging, requiring students to combine both equations with the reciprocal relationship and solve a system involving logarithms, but the approach is methodical rather than requiring deep insight. This is moderately above average difficulty for A-level, typical of AEA questions that test technical facility rather than novel problem-solving. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.06e Logarithm as inverse: ln(x) inverse of e^x1.06f Laws of logarithms: addition, subtraction, power rules |
3.Given that $x > y > 0$ ,
\begin{enumerate}[label=(\alph*)]
\item by writing $\log _ { y } x = z$ ,or otherwise,show that $\log _ { y } x = \frac { 1 } { \log _ { x } y }$ .
\item Given also that $\log _ { x } y = \log _ { y } x$ ,show that $y = \frac { 1 } { x }$ .
\item Solve the simultaneous equations
$$\begin{gathered}
\log _ { x } y = \log _ { y } x \\
\log _ { x } ( x - y ) = \log _ { y } ( x + y )
\end{gathered}$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel AEA 2006 Q3 [11]}}