Solve $LaTeX: \displaystyle \log_{ 4 }(x + 8) + \log_{ 4 }(x + 20) = 3$

Using the product rule for logarithms gives $LaTeX: \displaystyle \log_{ 4 }(\left(x + 8\right) \left(x + 20\right))$ and rewriting in exponential form gives $LaTeX: \displaystyle \left(x + 8\right) \left(x + 20\right) = 64$ expanding and setting the equation equal to zero gives $LaTeX: \displaystyle x^{2} + 28 x + 96 = 0$ . Factoring gives $LaTeX: \displaystyle \left(x + 4\right) \left(x + 24\right)=0$ . This gives two possible solutions $LaTeX: \displaystyle x=-24$ or $LaTeX: \displaystyle x=-4$ . $LaTeX: \displaystyle x=-24$ is an extraneous solution. The only soution is $LaTeX: \displaystyle x=-4$ .