Solve $LaTeX: \displaystyle \log_{ 6 }(x + 12) + \log_{ 6 }(x + 222) = 4$

Using the product rule for logarithms gives $LaTeX: \displaystyle \log_{ 6 }(\left(x + 12\right) \left(x + 222\right))$ and rewriting in exponential form gives $LaTeX: \displaystyle \left(x + 12\right) \left(x + 222\right) = 1296$ expanding and setting the equation equal to zero gives $LaTeX: \displaystyle x^{2} + 234 x + 1368 = 0$ . Factoring gives $LaTeX: \displaystyle \left(x + 6\right) \left(x + 228\right)=0$ . This gives two possible solutions $LaTeX: \displaystyle x=-228$ or $LaTeX: \displaystyle x=-6$ . $LaTeX: \displaystyle x=-228$ is an extraneous solution. The only soution is $LaTeX: \displaystyle x=-6$ .