Solve the equation $LaTeX: \displaystyle \log_{3}(x + 218)-\log_{3}(x + 56)=1$ .

Using the quotient property of logarithms gives $LaTeX: \displaystyle \log_{3}\frac{x + 218}{x + 56} = 1$ . Making both sides of the equation exponents on the base $LaTeX: \displaystyle 3$ gives $LaTeX: \displaystyle \frac{x + 218}{x + 56}=3$ . Clearing the fractions by multiplying by the LCD gives $LaTeX: \displaystyle x + 218=3 x + 168$ . Isolating $LaTeX: \displaystyle x$ gives $LaTeX: \displaystyle x = 25$ .