Solve $LaTeX: \displaystyle \log_{6}(x + 28)+\log_{6}(x + 239) = 5$ .

Using logarithmic properties and expanding the argument gives $LaTeX: \displaystyle \log_{6}(x^{2} + 267 x + 6692)=5$ . Making both sides an exponent on the base gives $LaTeX: \displaystyle x^{2} + 267 x + 6692=6^{5}$ . Expanding and setting equal to zero gives $LaTeX: \displaystyle x^{2} + 267 x - 1084=0$ . Factoring gives $LaTeX: \displaystyle \left(x - 4\right) \left(x + 271\right)=0$ . Solving gives the two possible solutions $LaTeX: \displaystyle x = -271$ and $LaTeX: \displaystyle x = 4$ . The domain of the original is $LaTeX: \displaystyle \left(-28, \infty\right) \bigcap \left(-239, \infty\right)=\left(-28, \infty\right)$ . Checking if each possible solution is in the domain gives: $LaTeX: \displaystyle x = -271$ is not a solution. $LaTeX: \displaystyle x=4$ is a solution.