Solve $LaTeX: \displaystyle \log_{ 12 }(x + 18) + \log_{ 12 }(x + 150) = 3$

Using the product rule for logarithms gives $LaTeX: \displaystyle \log_{ 12 }(\left(x + 18\right) \left(x + 150\right))$ and rewriting in exponential form gives $LaTeX: \displaystyle \left(x + 18\right) \left(x + 150\right) = 1728$ expanding and setting the equation equal to zero gives $LaTeX: \displaystyle x^{2} + 168 x + 972 = 0$ . Factoring gives $LaTeX: \displaystyle \left(x + 6\right) \left(x + 162\right)=0$ . This gives two possible solutions $LaTeX: \displaystyle x=-162$ or $LaTeX: \displaystyle x=-6$ . $LaTeX: \displaystyle x=-162$ is an extraneous solution. The only soution is $LaTeX: \displaystyle x=-6$ .