Solve $LaTeX: \displaystyle \log_{ 4 }(x + 6) + \log_{ 4 }(x + 6) = 2$

Using the product rule for logarithms gives $LaTeX: \displaystyle \log_{ 4 }(\left(x + 6\right)^{2})$ and rewriting in exponential form gives $LaTeX: \displaystyle \left(x + 6\right)^{2} = 16$ expanding and setting the equation equal to zero gives $LaTeX: \displaystyle x^{2} + 12 x + 20 = 0$ . Factoring gives $LaTeX: \displaystyle \left(x + 2\right) \left(x + 10\right)=0$ . This gives two possible solutions $LaTeX: \displaystyle x=-10$ or $LaTeX: \displaystyle x=-2$ . $LaTeX: \displaystyle x=-10$ is an extraneous solution. The only soution is $LaTeX: \displaystyle x=-2$ .