Solve $LaTeX: \displaystyle \log_{8}(x + 13)+\log_{8}(x + 1) = 2$ .

Using logarithmic properties and expanding the argument gives $LaTeX: \displaystyle \log_{8}(x^{2} + 14 x + 13)=2$ . Making both sides an exponent on the base gives $LaTeX: \displaystyle x^{2} + 14 x + 13=8^{2}$ . Expanding and setting equal to zero gives $LaTeX: \displaystyle x^{2} + 14 x - 51=0$ . Factoring gives $LaTeX: \displaystyle \left(x - 3\right) \left(x + 17\right)=0$ . Solving gives the two possible solutions $LaTeX: \displaystyle x = -17$ and $LaTeX: \displaystyle x = 3$ . The domain of the original is $LaTeX: \displaystyle \left(-13, \infty\right) \bigcap \left(-1, \infty\right)=\left(-1, \infty\right)$ . Checking if each possible solution is in the domain gives: $LaTeX: \displaystyle x = -17$ is not a solution. $LaTeX: \displaystyle x=3$ is a solution.