Solve $LaTeX: \displaystyle \log_{20}(x + 3115)+\log_{20}(x + 1014) = 5$ .

Using logarithmic properties and expanding the argument gives $LaTeX: \displaystyle \log_{20}(x^{2} + 4129 x + 3158610)=5$ . Making both sides an exponent on the base gives $LaTeX: \displaystyle x^{2} + 4129 x + 3158610=20^{5}$ . Expanding and setting equal to zero gives $LaTeX: \displaystyle x^{2} + 4129 x - 41390=0$ . Factoring gives $LaTeX: \displaystyle \left(x - 10\right) \left(x + 4139\right)=0$ . Solving gives the two possible solutions $LaTeX: \displaystyle x = -4139$ and $LaTeX: \displaystyle x = 10$ . The domain of the original is $LaTeX: \displaystyle \left(-3115, \infty\right) \bigcap \left(-1014, \infty\right)=\left(-1014, \infty\right)$ . Checking if each possible solution is in the domain gives: $LaTeX: \displaystyle x = -4139$ is not a solution. $LaTeX: \displaystyle x=10$ is a solution.