Logarithms
Change of Base
MMTS_Full_Test_08
Grade 12
Question:
If $\log_{12}27=a$, then $\log_6 16=$
$2\left(\dfrac{3-a}{3+a}\right)$
$3\left(\dfrac{3-a}{3+a}\right)$
$4\left(\dfrac{3-a}{3+a}\right)$
$4\left(\dfrac{2-a}{2+a}\right)$
Step-by-Step Solution
Key Concept: Express $\log 2$ and $\log 3$ in terms of $a$; then compute $\log_6 16$
From $a$: $\frac{\log 2}{\log 3}=\frac{3-a}{a+3}\cdot\frac{1}{...}$. After simplification: $\log_6 16=4\cdot\frac{3-a}{3+a}$.
Correct Answer: 3