Solve $LaTeX: \displaystyle \log_{ 16 }(x + 20) + \log_{ 16 }(x + 4100) = 4$

Using the product rule for logarithms gives $LaTeX: \displaystyle \log_{ 16 }(\left(x + 20\right) \left(x + 4100\right))$ and rewriting in exponential form gives $LaTeX: \displaystyle \left(x + 20\right) \left(x + 4100\right) = 65536$ expanding and setting the equation equal to zero gives $LaTeX: \displaystyle x^{2} + 4120 x + 16464 = 0$ . Factoring gives $LaTeX: \displaystyle \left(x + 4\right) \left(x + 4116\right)=0$ . This gives two possible solutions $LaTeX: \displaystyle x=-4116$ or $LaTeX: \displaystyle x=-4$ . $LaTeX: \displaystyle x=-4116$ is an extraneous solution. The only soution is $LaTeX: \displaystyle x=-4$ .