Solve $LaTeX: \displaystyle \log_{8}(x + 2)+\log_{8}(x + 58) = 3$ .

Using logarithmic properties and expanding the argument gives $LaTeX: \displaystyle \log_{8}(x^{2} + 60 x + 116)=3$ . Making both sides an exponent on the base gives $LaTeX: \displaystyle x^{2} + 60 x + 116=8^{3}$ . Expanding and setting equal to zero gives $LaTeX: \displaystyle x^{2} + 60 x - 396=0$ . Factoring gives $LaTeX: \displaystyle \left(x - 6\right) \left(x + 66\right)=0$ . Solving gives the two possible solutions $LaTeX: \displaystyle x = -66$ and $LaTeX: \displaystyle x = 6$ . The domain of the original is $LaTeX: \displaystyle \left(-2, \infty\right) \bigcap \left(-58, \infty\right)=\left(-2, \infty\right)$ . Checking if each possible solution is in the domain gives: $LaTeX: \displaystyle x = -66$ is not a solution. $LaTeX: \displaystyle x=6$ is a solution.