Solve $LaTeX: \displaystyle \log_{ 12 }(x + 15) + \log_{ 12 }(x + 147) = 3$

Using the product rule for logarithms gives $LaTeX: \displaystyle \log_{ 12 }(\left(x + 15\right) \left(x + 147\right))$ and rewriting in exponential form gives $LaTeX: \displaystyle \left(x + 15\right) \left(x + 147\right) = 1728$ expanding and setting the equation equal to zero gives $LaTeX: \displaystyle x^{2} + 162 x + 477 = 0$ . Factoring gives $LaTeX: \displaystyle \left(x + 3\right) \left(x + 159\right)=0$ . This gives two possible solutions $LaTeX: \displaystyle x=-159$ or $LaTeX: \displaystyle x=-3$ . $LaTeX: \displaystyle x=-159$ is an extraneous solution. The only soution is $LaTeX: \displaystyle x=-3$ .