Basic Maths and Logarithm
Properties of Logarithms
Premium Question
Grade 11
Question:
If $\log_{12} 27 = a$, then $\log_6 16$ equals:
(1) $\frac{4(3 - a)}{3 + a}$
(2) $\frac{4a}{3 - a}$
(3) $\frac{3 - a}{4}$
(4) $\frac{4(3 + a)}{3 - a}$
Step-by-Step Solution
Key Concept: $\log_{12} 27 = a \Rightarrow 3 \log 3 / (2 \log 2 + \log 3) = a$. Let $r = \log 2 / \log 3$: $3/(2r + 1) = a \Rightarrow r = (3 - a)/(2a)$. $\log_6 16 = 4 \log 2 / (\log 2 + \log 3) = 4/(1 + 1/r) = 4r / (r + 1) = \frac{4(3 - a)}{2a} \div \frac{(3 - a) + 2a}{2a} = \frac{4(3 - a)}{3 + a}$.
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Correct Answer: (1)