Solve the equation for \( x \). Give an exact solution if possible otherwise give an approximation to 3-decimal places. \( \log _{6}(x)-\log _{6}(x+4)=\log _{6}(x-3) \) \( x=\square \) Submit Question

To solve the equation \( \log _{6}(x) - \log _{6}(x+4) = \log _{6}(x-3) \), we can use the properties of logarithms. First, apply the quotient rule of logarithms: \[ \log_{6}\left(\frac{x}{x+4}\right) = \log_{6}(x-3) \] Now, since the bases are the same, we can set the arguments equal to each other: \[ \frac{x}{x+4} = x - 3 \] Cross-multiplying gives: \[ x = (x - 3)(x + 4) \] Expanding the right-hand side: \[ x = x^2 + 4x - 3x - 12 \implies x = x^2 + x - 12 \] Rearranging this equation results in: \[ 0 = x^2 - 12 \] Factoring gives: \[ 0 = (x - 4)(x + 3) \] Setting each factor to zero gives the solutions \( x = 4 \) and \( x = -3 \). However, we must check if both solutions are valid in the context of the logarithm. Since logarithms are only defined for positive arguments: - For \( x = 4 \): - \( \log_{6}(4) \), \( \log_{6}(8) \), and \( \log_{6}(1) \) are defined. - For \( x = -3 \): - \( \log_{6}(-3) \), \( \log_{6}(1) \), and \( \log_{6}(1) \) are not defined. Thus, the only valid solution is: \[ x = 4 \]