Solve $LaTeX: \displaystyle \log_{8}(x - 2)+\log_{8}(x + 10) = 2$ .

Using logarithmic properties and expanding the argument gives $LaTeX: \displaystyle \log_{8}(x^{2} + 8 x - 20)=2$ . Making both sides an exponent on the base gives $LaTeX: \displaystyle x^{2} + 8 x - 20=8^{2}$ . Expanding and setting equal to zero gives $LaTeX: \displaystyle x^{2} + 8 x - 84=0$ . Factoring gives $LaTeX: \displaystyle \left(x - 6\right) \left(x + 14\right)=0$ . Solving gives the two possible solutions $LaTeX: \displaystyle x = -14$ and $LaTeX: \displaystyle x = 6$ . The domain of the original is $LaTeX: \displaystyle \left(2, \infty\right) \bigcap \left(-10, \infty\right)=\left(2, \infty\right)$ . Checking if each possible solution is in the domain gives: $LaTeX: \displaystyle x = -14$ is not a solution. $LaTeX: \displaystyle x=6$ is a solution.