To find the upper bound approximation of \( \log_{11}(10) \) to four decimal places, we first calculate \( \log_{11}(10) \).
Using the change of base formula: \[ \log_{11}(10) = \frac{\log_{10}(10)}{\log_{10}(11)} \] Since \( \log_{10}(10) = 1 \), we need to find \( \log_{10}(11) \).
Calculating \( \log_{10}(11) \) using approximations (or a calculator), we find: \[ \log_{10}(11) \approx 1.041393 \] So: \[ \log_{11}(10) = \frac{1}{\log_{10}(11)} \approx \frac{1}{1.041393} \approx 0.960 \]
Thus, we need to take the logarithm value \( 1.041393 \) for comparison with the provided options. To round it to four decimal places, we have:
1.041393 rounds to 1.0414.
Therefore, the upper bound approximation is:
1.04140