Solve $LaTeX: \displaystyle \log_{ 3 }(x + 6) + \log_{ 3 }(x + 12) = 3$

Using the product rule for logarithms gives $LaTeX: \displaystyle \log_{ 3 }(\left(x + 6\right) \left(x + 12\right))$ and rewriting in exponential form gives $LaTeX: \displaystyle \left(x + 6\right) \left(x + 12\right) = 27$ expanding and setting the equation equal to zero gives $LaTeX: \displaystyle x^{2} + 18 x + 45 = 0$ . Factoring gives $LaTeX: \displaystyle \left(x + 3\right) \left(x + 15\right)=0$ . This gives two possible solutions $LaTeX: \displaystyle x=-15$ or $LaTeX: \displaystyle x=-3$ . $LaTeX: \displaystyle x=-15$ is an extraneous solution. The only soution is $LaTeX: \displaystyle x=-3$ .