Solve $LaTeX: \displaystyle \log_{ 12 }(x + 16) + \log_{ 12 }(x + 16) = 2$

Using the product rule for logarithms gives $LaTeX: \displaystyle \log_{ 12 }(\left(x + 16\right)^{2})$ and rewriting in exponential form gives $LaTeX: \displaystyle \left(x + 16\right)^{2} = 144$ expanding and setting the equation equal to zero gives $LaTeX: \displaystyle x^{2} + 32 x + 112 = 0$ . Factoring gives $LaTeX: \displaystyle \left(x + 4\right) \left(x + 28\right)=0$ . This gives two possible solutions $LaTeX: \displaystyle x=-28$ or $LaTeX: \displaystyle x=-4$ . $LaTeX: \displaystyle x=-28$ is an extraneous solution. The only soution is $LaTeX: \displaystyle x=-4$ .