Solve $LaTeX: \displaystyle \log_{15}(x + 122)+\log_{15}(x + 24) = 3$ .

Using logarithmic properties and expanding the argument gives $LaTeX: \displaystyle \log_{15}(x^{2} + 146 x + 2928)=3$ . Making both sides an exponent on the base gives $LaTeX: \displaystyle x^{2} + 146 x + 2928=15^{3}$ . Expanding and setting equal to zero gives $LaTeX: \displaystyle x^{2} + 146 x - 447=0$ . Factoring gives $LaTeX: \displaystyle \left(x - 3\right) \left(x + 149\right)=0$ . Solving gives the two possible solutions $LaTeX: \displaystyle x = -149$ and $LaTeX: \displaystyle x = 3$ . The domain of the original is $LaTeX: \displaystyle \left(-122, \infty\right) \bigcap \left(-24, \infty\right)=\left(-24, \infty\right)$ . Checking if each possible solution is in the domain gives: $LaTeX: \displaystyle x = -149$ is not a solution. $LaTeX: \displaystyle x=3$ is a solution.