Solve $LaTeX: \displaystyle \log_{8}(x)+\log_{8}(x + 12) = 2$ .

Using logarithmic properties and expanding the argument gives $LaTeX: \displaystyle \log_{8}(x^{2} + 12 x)=2$ . Making both sides an exponent on the base gives $LaTeX: \displaystyle x^{2} + 12 x=8^{2}$ . Expanding and setting equal to zero gives $LaTeX: \displaystyle x^{2} + 12 x - 64=0$ . Factoring gives $LaTeX: \displaystyle \left(x - 4\right) \left(x + 16\right)=0$ . Solving gives the two possible solutions $LaTeX: \displaystyle x = -16$ and $LaTeX: \displaystyle x = 4$ . The domain of the original is $LaTeX: \displaystyle \left(0, \infty\right) \bigcap \left(-12, \infty\right)=\left(0, \infty\right)$ . Checking if each possible solution is in the domain gives: $LaTeX: \displaystyle x = -16$ is not a solution. $LaTeX: \displaystyle x=4$ is a solution.