Solve $LaTeX: \displaystyle \log_{ 6 }(x + 8) + \log_{ 6 }(x + 38) = 3$

Using the product rule for logarithms gives $LaTeX: \displaystyle \log_{ 6 }(\left(x + 8\right) \left(x + 38\right))$ and rewriting in exponential form gives $LaTeX: \displaystyle \left(x + 8\right) \left(x + 38\right) = 216$ expanding and setting the equation equal to zero gives $LaTeX: \displaystyle x^{2} + 46 x + 88 = 0$ . Factoring gives $LaTeX: \displaystyle \left(x + 2\right) \left(x + 44\right)=0$ . This gives two possible solutions $LaTeX: \displaystyle x=-44$ or $LaTeX: \displaystyle x=-2$ . $LaTeX: \displaystyle x=-44$ is an extraneous solution. The only soution is $LaTeX: \displaystyle x=-2$ .