Solve $LaTeX: \displaystyle \log_{ 16 }(x + 20) + \log_{ 16 }(x + 20) = 2$

Using the product rule for logarithms gives $LaTeX: \displaystyle \log_{ 16 }(\left(x + 20\right)^{2})$ and rewriting in exponential form gives $LaTeX: \displaystyle \left(x + 20\right)^{2} = 256$ expanding and setting the equation equal to zero gives $LaTeX: \displaystyle x^{2} + 40 x + 144 = 0$ . Factoring gives $LaTeX: \displaystyle \left(x + 4\right) \left(x + 36\right)=0$ . This gives two possible solutions $LaTeX: \displaystyle x=-36$ or $LaTeX: \displaystyle x=-4$ . $LaTeX: \displaystyle x=-36$ is an extraneous solution. The only soution is $LaTeX: \displaystyle x=-4$ .