Solve $LaTeX: \displaystyle \log_{ 18 }(x + 24) + \log_{ 18 }(x + 330) = 3$

Using the product rule for logarithms gives $LaTeX: \displaystyle \log_{ 18 }(\left(x + 24\right) \left(x + 330\right))$ and rewriting in exponential form gives $LaTeX: \displaystyle \left(x + 24\right) \left(x + 330\right) = 5832$ expanding and setting the equation equal to zero gives $LaTeX: \displaystyle x^{2} + 354 x + 2088 = 0$ . Factoring gives $LaTeX: \displaystyle \left(x + 6\right) \left(x + 348\right)=0$ . This gives two possible solutions $LaTeX: \displaystyle x=-348$ or $LaTeX: \displaystyle x=-6$ . $LaTeX: \displaystyle x=-348$ is an extraneous solution. The only soution is $LaTeX: \displaystyle x=-6$ .