Solve $LaTeX: \displaystyle \log_{ 12 }(x + 16) + \log_{ 12 }(x + 148) = 3$

Using the product rule for logarithms gives $LaTeX: \displaystyle \log_{ 12 }(\left(x + 16\right) \left(x + 148\right))$ and rewriting in exponential form gives $LaTeX: \displaystyle \left(x + 16\right) \left(x + 148\right) = 1728$ expanding and setting the equation equal to zero gives $LaTeX: \displaystyle x^{2} + 164 x + 640 = 0$ . Factoring gives $LaTeX: \displaystyle \left(x + 4\right) \left(x + 160\right)=0$ . This gives two possible solutions $LaTeX: \displaystyle x=-160$ or $LaTeX: \displaystyle x=-4$ . $LaTeX: \displaystyle x=-160$ is an extraneous solution. The only soution is $LaTeX: \displaystyle x=-4$ .