Solve $LaTeX: \displaystyle \log_{15}(x + 20)+\log_{15}(x + 4) = 2$ .

Using logarithmic properties and expanding the argument gives $LaTeX: \displaystyle \log_{15}(x^{2} + 24 x + 80)=2$ . Making both sides an exponent on the base gives $LaTeX: \displaystyle x^{2} + 24 x + 80=15^{2}$ . Expanding and setting equal to zero gives $LaTeX: \displaystyle x^{2} + 24 x - 145=0$ . Factoring gives $LaTeX: \displaystyle \left(x - 5\right) \left(x + 29\right)=0$ . Solving gives the two possible solutions $LaTeX: \displaystyle x = -29$ and $LaTeX: \displaystyle x = 5$ . The domain of the original is $LaTeX: \displaystyle \left(-20, \infty\right) \bigcap \left(-4, \infty\right)=\left(-4, \infty\right)$ . Checking if each possible solution is in the domain gives: $LaTeX: \displaystyle x = -29$ is not a solution. $LaTeX: \displaystyle x=5$ is a solution.