Solve $LaTeX: \displaystyle \log_{6}(x + 25)+\log_{6}(x + 6) = 3$ .

Using logarithmic properties and expanding the argument gives $LaTeX: \displaystyle \log_{6}(x^{2} + 31 x + 150)=3$ . Making both sides an exponent on the base gives $LaTeX: \displaystyle x^{2} + 31 x + 150=6^{3}$ . Expanding and setting equal to zero gives $LaTeX: \displaystyle x^{2} + 31 x - 66=0$ . Factoring gives $LaTeX: \displaystyle \left(x - 2\right) \left(x + 33\right)=0$ . Solving gives the two possible solutions $LaTeX: \displaystyle x = -33$ and $LaTeX: \displaystyle x = 2$ . The domain of the original is $LaTeX: \displaystyle \left(-25, \infty\right) \bigcap \left(-6, \infty\right)=\left(-6, \infty\right)$ . Checking if each possible solution is in the domain gives: $LaTeX: \displaystyle x = -33$ is not a solution. $LaTeX: \displaystyle x=2$ is a solution.