Solve $LaTeX: \displaystyle \log_{ 8 }(x + 12) + \log_{ 8 }(x + 68) = 3$

Using the product rule for logarithms gives $LaTeX: \displaystyle \log_{ 8 }(\left(x + 12\right) \left(x + 68\right))$ and rewriting in exponential form gives $LaTeX: \displaystyle \left(x + 12\right) \left(x + 68\right) = 512$ expanding and setting the equation equal to zero gives $LaTeX: \displaystyle x^{2} + 80 x + 304 = 0$ . Factoring gives $LaTeX: \displaystyle \left(x + 4\right) \left(x + 76\right)=0$ . This gives two possible solutions $LaTeX: \displaystyle x=-76$ or $LaTeX: \displaystyle x=-4$ . $LaTeX: \displaystyle x=-76$ is an extraneous solution. The only soution is $LaTeX: \displaystyle x=-4$ .