Edexcel AEA 2003 June — Question 6 19 marks

Exam BoardEdexcel
ModuleAEA (Advanced Extension Award)
Year2003
SessionJune
Marks19
PaperDownload PDF ↗
TopicLaws of Logarithms
TypeSimplify or prove logarithmic identity
DifficultyChallenging +1.8 Part (a) requires algebraic manipulation by squaring both sides—a non-routine technique. Part (b) is straightforward application once (a) is done. Part (c) demands systematic problem-solving to find integer solutions, requiring insight into nested radicals and logarithm properties. This is AEA material requiring extended reasoning beyond standard A-level, but the techniques are accessible to strong students.
Spec1.02b Surds: manipulation and rationalising denominators1.06f Laws of logarithms: addition, subtraction, power rules
6.(a)Show that $$\sqrt { 2 + \sqrt { 3 } } - \sqrt { 2 - \sqrt { 3 } } = \sqrt { 2 }$$ (b)Hence prove that $$\log _ { \frac { 1 } { 8 } } ( \sqrt { 2 + \sqrt { 3 } } - \sqrt { 2 - \sqrt { 3 } } ) = - \frac { 1 } { 6 } .$$ (c)Find all possible pairs of integers \(a\) and \(n\) such that $$\log _ { \frac { 1 } { n } } ( \sqrt { a + \sqrt { 15 } } - \sqrt { a - \sqrt { 15 } } ) = - \frac { 1 } { 2 } .$$ 
6.(a)Show that

$$\sqrt { 2 + \sqrt { 3 } } - \sqrt { 2 - \sqrt { 3 } } = \sqrt { 2 }$$

(b)Hence prove that

$$\log _ { \frac { 1 } { 8 } } ( \sqrt { 2 + \sqrt { 3 } } - \sqrt { 2 - \sqrt { 3 } } ) = - \frac { 1 } { 6 } .$$

(c)Find all possible pairs of integers $a$ and $n$ such that

$$\log _ { \frac { 1 } { n } } ( \sqrt { a + \sqrt { 15 } } - \sqrt { a - \sqrt { 15 } } ) = - \frac { 1 } { 2 } .$$

\hfill \mbox{\textit{Edexcel AEA 2003 Q6 [19]}}
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Q1 5 Q2 8 Q3 11 Q4 11 Q5 17 Q6 19 Q7 22