Evaluate the following expression using the properties of logarithms. \( \log _{6}(6)+\log _{6}(36)+\log _{6}(216) \) 36 43 13

To evaluate \( \log_{6}(6) + \log_{6}(36) + \log_{6}(216) \), we can use the properties of logarithms. First, \( \log_{6}(6) = 1 \) because any log base \( b \) of \( b \) is \( 1 \). Next, we can express \( 36 \) and \( 216 \) in terms of powers of \( 6 \): - \( 36 = 6^2 \), so \( \log_{6}(36) = 2 \). - \( 216 = 6^3 \), so \( \log_{6}(216) = 3 \). Now, we can combine these results: \[ \log_{6}(6) + \log_{6}(36) + \log_{6}(216) = 1 + 2 + 3 = 6. \] Thus, the overall result is 6. It appears that there might be a misunderstanding in the provided options (36, 43, 13), as the sum totals to 6 instead!