Solve $LaTeX: \displaystyle \log_{15}(x + 115)+\log_{15}(x + 17) = 3$ .

Using logarithmic properties and expanding the argument gives $LaTeX: \displaystyle \log_{15}(x^{2} + 132 x + 1955)=3$ . Making both sides an exponent on the base gives $LaTeX: \displaystyle x^{2} + 132 x + 1955=15^{3}$ . Expanding and setting equal to zero gives $LaTeX: \displaystyle x^{2} + 132 x - 1420=0$ . Factoring gives $LaTeX: \displaystyle \left(x - 10\right) \left(x + 142\right)=0$ . Solving gives the two possible solutions $LaTeX: \displaystyle x = -142$ and $LaTeX: \displaystyle x = 10$ . The domain of the original is $LaTeX: \displaystyle \left(-115, \infty\right) \bigcap \left(-17, \infty\right)=\left(-17, \infty\right)$ . Checking if each possible solution is in the domain gives: $LaTeX: \displaystyle x = -142$ is not a solution. $LaTeX: \displaystyle x=10$ is a solution.