Solve $LaTeX: \displaystyle \log_{ 4 }(x + 8) + \log_{ 4 }(x + 68) = 4$

Using the product rule for logarithms gives $LaTeX: \displaystyle \log_{ 4 }(\left(x + 8\right) \left(x + 68\right))$ and rewriting in exponential form gives $LaTeX: \displaystyle \left(x + 8\right) \left(x + 68\right) = 256$ expanding and setting the equation equal to zero gives $LaTeX: \displaystyle x^{2} + 76 x + 288 = 0$ . Factoring gives $LaTeX: \displaystyle \left(x + 4\right) \left(x + 72\right)=0$ . This gives two possible solutions $LaTeX: \displaystyle x=-72$ or $LaTeX: \displaystyle x=-4$ . $LaTeX: \displaystyle x=-72$ is an extraneous solution. The only soution is $LaTeX: \displaystyle x=-4$ .