Solve $LaTeX: \displaystyle \log_{ 18 }(x + 24) + \log_{ 18 }(x + 24) = 2$

Using the product rule for logarithms gives $LaTeX: \displaystyle \log_{ 18 }(\left(x + 24\right)^{2})$ and rewriting in exponential form gives $LaTeX: \displaystyle \left(x + 24\right)^{2} = 324$ expanding and setting the equation equal to zero gives $LaTeX: \displaystyle x^{2} + 48 x + 252 = 0$ . Factoring gives $LaTeX: \displaystyle \left(x + 6\right) \left(x + 42\right)=0$ . This gives two possible solutions $LaTeX: \displaystyle x=-42$ or $LaTeX: \displaystyle x=-6$ . $LaTeX: \displaystyle x=-42$ is an extraneous solution. The only soution is $LaTeX: \displaystyle x=-6$ .