1. The problem is to compute $\log_6 8$ using the change of base formula and round the answer to the nearest thousandth. 2. The change of base formula states: $$\log_a b = \frac{\log_c b}{\log_c a}$$ where $a$ and $b$ are positive numbers, $a \neq 1$, and $c$ is any positive base (commonly 10 or $e$). 3. Using base 10 logarithms, we write: $$\log_6 8 = \frac{\log_{10} 8}{\log_{10} 6}$$ 4. Calculate the numerator and denominator: $$\log_{10} 8 \approx 0.9031$$ $$\log_{10} 6 \approx 0.7782$$ 5. Divide the two values: $$\log_6 8 = \frac{0.9031}{0.7782}$$ 6. Show cancellation step: $$\frac{\cancel{0.9031}}{\cancel{0.7782}} = 1.160$$ 7. Therefore, rounded to the nearest thousandth: $$\log_6 8 \approx 1.160$$ This means that $6^{1.160} \approx 8$.