Use the change of base formula to compute \( \log _{6} \frac{1}{8} \). Round your answer to the nearest thousandth.

To compute \( \log_{6} \frac{1}{8} \) using the change of base formula, we can express the logarithm in terms of the natural logarithm (ln) or the common logarithm (log). The change of base formula is: \[ \log_{b} a = \frac{\log_{c} a}{\log_{c} b} \] where \( c \) is the base of the new logarithm. We can choose \( c \) to be either \( e \) (the base of the natural logarithm) or 10 (the base of the common logarithm). For this problem, let's use the natural logarithm (ln) and the base \( e \): \[ \log_{6} \frac{1}{8} = \frac{\ln \left( \frac{1}{8} \right)}{\ln 6} \] Now, let's calculate the natural logarithms of \( \frac{1}{8} \) and 6: \[ \ln \left( \frac{1}{8} \right) = \ln \left( 8^{-1} \right) = -\ln 8 \] \[ \ln 8 = \ln (2^3) = 3 \ln 2 \] We can use the known value of \( \ln 2 \) which is approximately 0.69314718056. Therefore: \[ \ln 8 = 3 \times 0.69314718056 \approx 2.07944154268 \] So, \[ \ln \left( \frac{1}{8} \right) = -2.07944154268 \] Now, let's calculate \( \ln 6 \): \[ \ln 6 = \ln (2 \times 3) = \ln 2 + \ln 3 \] We already know \( \ln 2 \approx 0.69314718056 \), and \( \ln 3 \) is approximately 1.09861228866. Therefore: \[ \ln 6 = 0.69314718056 + 1.09861228866 \approx 1.79175946922 \] Now we can put it all together: \[ \log_{6} \frac{1}{8} = \frac{-2.07944154268}{1.79175946922} \approx -1.161 \] Rounded to the nearest thousandth, the answer is: \[ \log_{6} \frac{1}{8} \approx -1.161 \]