Solve $LaTeX: \displaystyle \log_{ 10 }(x + 12) + \log_{ 10 }(x + 102) = 3$

Using the product rule for logarithms gives $LaTeX: \displaystyle \log_{ 10 }(\left(x + 12\right) \left(x + 102\right))$ and rewriting in exponential form gives $LaTeX: \displaystyle \left(x + 12\right) \left(x + 102\right) = 1000$ expanding and setting the equation equal to zero gives $LaTeX: \displaystyle x^{2} + 114 x + 224 = 0$ . Factoring gives $LaTeX: \displaystyle \left(x + 2\right) \left(x + 112\right)=0$ . This gives two possible solutions $LaTeX: \displaystyle x=-112$ or $LaTeX: \displaystyle x=-2$ . $LaTeX: \displaystyle x=-112$ is an extraneous solution. The only soution is $LaTeX: \displaystyle x=-2$ .