Solve $LaTeX: \displaystyle \log_{15}(x + 19)+\log_{15}(x + 3) = 2$ .

Using logarithmic properties and expanding the argument gives $LaTeX: \displaystyle \log_{15}(x^{2} + 22 x + 57)=2$ . Making both sides an exponent on the base gives $LaTeX: \displaystyle x^{2} + 22 x + 57=15^{2}$ . Expanding and setting equal to zero gives $LaTeX: \displaystyle x^{2} + 22 x - 168=0$ . Factoring gives $LaTeX: \displaystyle \left(x - 6\right) \left(x + 28\right)=0$ . Solving gives the two possible solutions $LaTeX: \displaystyle x = -28$ and $LaTeX: \displaystyle x = 6$ . The domain of the original is $LaTeX: \displaystyle \left(-19, \infty\right) \bigcap \left(-3, \infty\right)=\left(-3, \infty\right)$ . Checking if each possible solution is in the domain gives: $LaTeX: \displaystyle x = -28$ is not a solution. $LaTeX: \displaystyle x=6$ is a solution.