Solve $LaTeX: \displaystyle \log_{8}(x + 60)+\log_{8}(x + 4) = 3$ .

Using logarithmic properties and expanding the argument gives $LaTeX: \displaystyle \log_{8}(x^{2} + 64 x + 240)=3$ . Making both sides an exponent on the base gives $LaTeX: \displaystyle x^{2} + 64 x + 240=8^{3}$ . Expanding and setting equal to zero gives $LaTeX: \displaystyle x^{2} + 64 x - 272=0$ . Factoring gives $LaTeX: \displaystyle \left(x - 4\right) \left(x + 68\right)=0$ . Solving gives the two possible solutions $LaTeX: \displaystyle x = -68$ and $LaTeX: \displaystyle x = 4$ . The domain of the original is $LaTeX: \displaystyle \left(-60, \infty\right) \bigcap \left(-4, \infty\right)=\left(-4, \infty\right)$ . Checking if each possible solution is in the domain gives: $LaTeX: \displaystyle x = -68$ is not a solution. $LaTeX: \displaystyle x=4$ is a solution.