Solve $LaTeX: \displaystyle \log_{ 10 }(x + 12) + \log_{ 10 }(x + 1002) = 4$

Using the product rule for logarithms gives $LaTeX: \displaystyle \log_{ 10 }(\left(x + 12\right) \left(x + 1002\right))$ and rewriting in exponential form gives $LaTeX: \displaystyle \left(x + 12\right) \left(x + 1002\right) = 10000$ expanding and setting the equation equal to zero gives $LaTeX: \displaystyle x^{2} + 1014 x + 2024 = 0$ . Factoring gives $LaTeX: \displaystyle \left(x + 2\right) \left(x + 1012\right)=0$ . This gives two possible solutions $LaTeX: \displaystyle x=-1012$ or $LaTeX: \displaystyle x=-2$ . $LaTeX: \displaystyle x=-1012$ is an extraneous solution. The only soution is $LaTeX: \displaystyle x=-2$ .