Solve $LaTeX: \displaystyle \log_{15}(x + 123)+\log_{15}(x + 25) = 3$ .

Using logarithmic properties and expanding the argument gives $LaTeX: \displaystyle \log_{15}(x^{2} + 148 x + 3075)=3$ . Making both sides an exponent on the base gives $LaTeX: \displaystyle x^{2} + 148 x + 3075=15^{3}$ . Expanding and setting equal to zero gives $LaTeX: \displaystyle x^{2} + 148 x - 300=0$ . Factoring gives $LaTeX: \displaystyle \left(x - 2\right) \left(x + 150\right)=0$ . Solving gives the two possible solutions $LaTeX: \displaystyle x = -150$ and $LaTeX: \displaystyle x = 2$ . The domain of the original is $LaTeX: \displaystyle \left(-123, \infty\right) \bigcap \left(-25, \infty\right)=\left(-25, \infty\right)$ . Checking if each possible solution is in the domain gives: $LaTeX: \displaystyle x = -150$ is not a solution. $LaTeX: \displaystyle x=2$ is a solution.