Solve $LaTeX: \displaystyle \log_{ 6 }(x + 12) + \log_{ 6 }(x + 42) = 3$

Using the product rule for logarithms gives $LaTeX: \displaystyle \log_{ 6 }(\left(x + 12\right) \left(x + 42\right))$ and rewriting in exponential form gives $LaTeX: \displaystyle \left(x + 12\right) \left(x + 42\right) = 216$ expanding and setting the equation equal to zero gives $LaTeX: \displaystyle x^{2} + 54 x + 288 = 0$ . Factoring gives $LaTeX: \displaystyle \left(x + 6\right) \left(x + 48\right)=0$ . This gives two possible solutions $LaTeX: \displaystyle x=-48$ or $LaTeX: \displaystyle x=-6$ . $LaTeX: \displaystyle x=-48$ is an extraneous solution. The only soution is $LaTeX: \displaystyle x=-6$ .