Solve $LaTeX: \displaystyle \log_{12}(x - 1)+\log_{12}(x + 6) = 2$ .

Using logarithmic properties and expanding the argument gives $LaTeX: \displaystyle \log_{12}(x^{2} + 5 x - 6)=2$ . Making both sides an exponent on the base gives $LaTeX: \displaystyle x^{2} + 5 x - 6=12^{2}$ . Expanding and setting equal to zero gives $LaTeX: \displaystyle x^{2} + 5 x - 150=0$ . Factoring gives $LaTeX: \displaystyle \left(x - 10\right) \left(x + 15\right)=0$ . Solving gives the two possible solutions $LaTeX: \displaystyle x = -15$ and $LaTeX: \displaystyle x = 10$ . The domain of the original is $LaTeX: \displaystyle \left(1, \infty\right) \bigcap \left(-6, \infty\right)=\left(1, \infty\right)$ . Checking if each possible solution is in the domain gives: $LaTeX: \displaystyle x = -15$ is not a solution. $LaTeX: \displaystyle x=10$ is a solution.