Solve the equation. \( \begin{array}{c}\log _{7}(5 x-1)=\log _{7}(x+7) \\ x=[?]\end{array} \)

To solve the equation \( \log_{7}(5x - 1) = \log_{7}(x + 7) \), we can utilize the property that if \( \log_{b}(A) = \log_{b}(B) \), then \( A = B \). Thus, we can set the arguments equal to each other: \[ 5x - 1 = x + 7. \] Now, we'll solve for \( x \). First, subtract \( x \) from both sides: \[ 5x - x - 1 = 7, \] which simplifies to: \[ 4x - 1 = 7. \] Next, add \( 1 \) to both sides: \[ 4x = 8. \] Now, divide by \( 4 \): \[ x = 2. \] Finally, it's always good to check our solution by substituting it back into the original equation. For \( x = 2 \): \[ \log_{7}(5(2) - 1) = \log_{7}(2 + 7) \\ \log_{7}(10 - 1) = \log_{7}(9) \\ \log_{7}(9) = \log_{7}(9). \] Both sides match, so our solution is valid. Thus, the solution is: \[ x = 2. \]