Question
Consider $a=\log _{63} 64 \times \log _{62} 63 \times \log _{61} 62 \times \cdots \times \log _2 3$. Given that $a \in \mathbb{Z}$, find the value of $a$.
Answer/Explanation
Using the properties of logarithms, we get
$
\begin{aligned}
a & =\log _{63} 64 \times \log _{62} 63 \times \log _{61} 62 \times \cdots \times \log _2 3 \\
& =\frac{\log 64}{\ log 63} \times \frac{\log 63}{\log 62} \times \frac{\log 62}{\log 61} \times \cdots \times \frac{\log 3}{\log 2} \\
& =\frac{\log 64}{\log 2} \\
& =\frac{\log 2^6}{\log 2} \\
& =\frac{6 \log 2}{\log 2} \\
& =6
\end{aligned}
$
Question
Consider the expansion of \(\left ( 8x^{3} -\frac{1}{2x} \right )^{n}\) where n ∈ R+. Determine all possible values of n for which the expansion has a non-zero constant term.
Answer/Explanation
Ans:
EITHER
attempt to obtain the general term of the expansion
OR
recognize power of x starts at 3n and goes down by 4 each time
THEN
recognizing the constant term when the power of x is zero (or equivalent)
\(r = \frac{3n}{4} or n = \frac{4}{3}r or 3n – 4r = 0 OR 3r – (n-r) = 0 (or equivalent)\)
r is a multiple of 3 (r = 3,6,9,…) or one correct value of n (seen anywhere)
n = 4k, k ∈ Z+
Note: Accept n is a (positive) multiple of 4 or n = 4,8,12,…
Do not accept n = 4,8,12
Note: Award full marks for a correct answer using trial and error approach showing n = 4,8,12,… and for recognizing that this pattern continues.