Solve $LaTeX: \displaystyle \log_{10}(x + 123)+\log_{10}(x + 6) = 3$ .

Using logarithmic properties and expanding the argument gives $LaTeX: \displaystyle \log_{10}(x^{2} + 129 x + 738)=3$ . Making both sides an exponent on the base gives $LaTeX: \displaystyle x^{2} + 129 x + 738=10^{3}$ . Expanding and setting equal to zero gives $LaTeX: \displaystyle x^{2} + 129 x - 262=0$ . Factoring gives $LaTeX: \displaystyle \left(x - 2\right) \left(x + 131\right)=0$ . Solving gives the two possible solutions $LaTeX: \displaystyle x = -131$ and $LaTeX: \displaystyle x = 2$ . The domain of the original is $LaTeX: \displaystyle \left(-123, \infty\right) \bigcap \left(-6, \infty\right)=\left(-6, \infty\right)$ . Checking if each possible solution is in the domain gives: $LaTeX: \displaystyle x = -131$ is not a solution. $LaTeX: \displaystyle x=2$ is a solution.