Solve for \( x \) \( \log _{6} x-\log _{6}(x+2)=\log _{6}\left(\frac{1}{6}\right) \) \( \square \)
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Let's simplify the equation step by step. Using the property of logarithms, we can combine the left side: \[ \log_{6} x - \log_{6}(x + 2) = \log_{6}\left(\frac{x}{x + 2}\right) \] So our equation becomes: \[ \log_{6}\left(\frac{x}{x + 2}\right) = \log_{6}\left(\frac{1}{6}\right) \] Since the logarithms are equal, we can set the insides equal to each other: \[ \frac{x}{x + 2} = \frac{1}{6} \] Cross-multiplying gives: \[ 6x = x + 2 \] Now, simplify and solve for \( x \): \[ 6x - x = 2 \Rightarrow 5x = 2 \Rightarrow x = \frac{2}{5} \] Thus, the solution is: \[ x = \frac{2}{5} \]