Solve $LaTeX: \displaystyle \log_{15}(x + 121)+\log_{15}(x + 23) = 3$ .

Using logarithmic properties and expanding the argument gives $LaTeX: \displaystyle \log_{15}(x^{2} + 144 x + 2783)=3$ . Making both sides an exponent on the base gives $LaTeX: \displaystyle x^{2} + 144 x + 2783=15^{3}$ . Expanding and setting equal to zero gives $LaTeX: \displaystyle x^{2} + 144 x - 592=0$ . Factoring gives $LaTeX: \displaystyle \left(x - 4\right) \left(x + 148\right)=0$ . Solving gives the two possible solutions $LaTeX: \displaystyle x = -148$ and $LaTeX: \displaystyle x = 4$ . The domain of the original is $LaTeX: \displaystyle \left(-121, \infty\right) \bigcap \left(-23, \infty\right)=\left(-23, \infty\right)$ . Checking if each possible solution is in the domain gives: $LaTeX: \displaystyle x = -148$ is not a solution. $LaTeX: \displaystyle x=4$ is a solution.