Solve $LaTeX: \displaystyle \log_{8}(x + 2)+\log_{8}(x + 14) = 2$ .

Using logarithmic properties and expanding the argument gives $LaTeX: \displaystyle \log_{8}(x^{2} + 16 x + 28)=2$ . Making both sides an exponent on the base gives $LaTeX: \displaystyle x^{2} + 16 x + 28=8^{2}$ . Expanding and setting equal to zero gives $LaTeX: \displaystyle x^{2} + 16 x - 36=0$ . Factoring gives $LaTeX: \displaystyle \left(x - 2\right) \left(x + 18\right)=0$ . Solving gives the two possible solutions $LaTeX: \displaystyle x = -18$ and $LaTeX: \displaystyle x = 2$ . The domain of the original is $LaTeX: \displaystyle \left(-2, \infty\right) \bigcap \left(-14, \infty\right)=\left(-2, \infty\right)$ . Checking if each possible solution is in the domain gives: $LaTeX: \displaystyle x = -18$ is not a solution. $LaTeX: \displaystyle x=2$ is a solution.