Solve the equation $LaTeX: \displaystyle \log_{6}(x + 1298)-\log_{6}(x + 218)=1$ .

Using the quotient property of logarithms gives $LaTeX: \displaystyle \log_{6}\frac{x + 1298}{x + 218} = 1$ . Making both sides of the equation exponents on the base $LaTeX: \displaystyle 6$ gives $LaTeX: \displaystyle \frac{x + 1298}{x + 218}=6$ . Clearing the fractions by multiplying by the LCD gives $LaTeX: \displaystyle x + 1298=6 x + 1308$ . Isolating $LaTeX: \displaystyle x$ gives $LaTeX: \displaystyle x = -2$ .