### Instruction: If $\log_{3}6 \approx 1.63$, what is the value of $\log_{9} 4$, accurate to two decimal places?### Response: I know that $\log_{9} 4$ is the same as $\frac{\log_{3} 4}{\log_{3} 9}$, by the change of base formula.
To find $\log_{3} 4$, I can use the fact that $4 = 2^2$, so $\log_{3} 4 = 2 \log_{3} 2$.
I also know that $\log_{3} 9 = 2$, since $9 = 3^2$.
Therefore, $\log_{9} 4 = \frac{2 \log_{3} 2}{2} = \log_{3} 2$.
Now, how can I find $\log_{3} 2$?
I notice that $6 = 2 \cdot 3$, so $\log_{3} 6 = \log_{3} 2 + \log_{3} 3$.
Since $\log_{3} 3 = 1$, I can solve for $\log_{3} 2$ by subtracting 1 from both sides: $\log_{3} 2 = \log_{3} 6 - 1$.
The problem gives me an approximate value for $\log_{3} 6$, which is 1.63.
Therefore, $\log_{3} 2 \approx 1.63 - 1 = 0.63$.