Solve $LaTeX: \displaystyle \log_{ 6 }(x + 9) + \log_{ 6 }(x + 39) = 3$

Using the product rule for logarithms gives $LaTeX: \displaystyle \log_{ 6 }(\left(x + 9\right) \left(x + 39\right))$ and rewriting in exponential form gives $LaTeX: \displaystyle \left(x + 9\right) \left(x + 39\right) = 216$ expanding and setting the equation equal to zero gives $LaTeX: \displaystyle x^{2} + 48 x + 135 = 0$ . Factoring gives $LaTeX: \displaystyle \left(x + 3\right) \left(x + 45\right)=0$ . This gives two possible solutions $LaTeX: \displaystyle x=-45$ or $LaTeX: \displaystyle x=-3$ . $LaTeX: \displaystyle x=-45$ is an extraneous solution. The only soution is $LaTeX: \displaystyle x=-3$ .