Solve $LaTeX: \displaystyle \log_{6}(x + 240)+\log_{6}(x + 29) = 5$ .

Using logarithmic properties and expanding the argument gives $LaTeX: \displaystyle \log_{6}(x^{2} + 269 x + 6960)=5$ . Making both sides an exponent on the base gives $LaTeX: \displaystyle x^{2} + 269 x + 6960=6^{5}$ . Expanding and setting equal to zero gives $LaTeX: \displaystyle x^{2} + 269 x - 816=0$ . Factoring gives $LaTeX: \displaystyle \left(x - 3\right) \left(x + 272\right)=0$ . Solving gives the two possible solutions $LaTeX: \displaystyle x = -272$ and $LaTeX: \displaystyle x = 3$ . The domain of the original is $LaTeX: \displaystyle \left(-240, \infty\right) \bigcap \left(-29, \infty\right)=\left(-29, \infty\right)$ . Checking if each possible solution is in the domain gives: $LaTeX: \displaystyle x = -272$ is not a solution. $LaTeX: \displaystyle x=3$ is a solution.