Solve $LaTeX: \displaystyle \log_{ 6 }(x + 9) + \log_{ 6 }(x + 219) = 4$

Using the product rule for logarithms gives $LaTeX: \displaystyle \log_{ 6 }(\left(x + 9\right) \left(x + 219\right))$ and rewriting in exponential form gives $LaTeX: \displaystyle \left(x + 9\right) \left(x + 219\right) = 1296$ expanding and setting the equation equal to zero gives $LaTeX: \displaystyle x^{2} + 228 x + 675 = 0$ . Factoring gives $LaTeX: \displaystyle \left(x + 3\right) \left(x + 225\right)=0$ . This gives two possible solutions $LaTeX: \displaystyle x=-225$ or $LaTeX: \displaystyle x=-3$ . $LaTeX: \displaystyle x=-225$ is an extraneous solution. The only soution is $LaTeX: \displaystyle x=-3$ .