Solve $LaTeX: \displaystyle \log_{10}(x + 118)+\log_{10}(x + 1) = 3$ .

Using logarithmic properties and expanding the argument gives $LaTeX: \displaystyle \log_{10}(x^{2} + 119 x + 118)=3$ . Making both sides an exponent on the base gives $LaTeX: \displaystyle x^{2} + 119 x + 118=10^{3}$ . Expanding and setting equal to zero gives $LaTeX: \displaystyle x^{2} + 119 x - 882=0$ . Factoring gives $LaTeX: \displaystyle \left(x - 7\right) \left(x + 126\right)=0$ . Solving gives the two possible solutions $LaTeX: \displaystyle x = -126$ and $LaTeX: \displaystyle x = 7$ . The domain of the original is $LaTeX: \displaystyle \left(-118, \infty\right) \bigcap \left(-1, \infty\right)=\left(-1, \infty\right)$ . Checking if each possible solution is in the domain gives: $LaTeX: \displaystyle x = -126$ is not a solution. $LaTeX: \displaystyle x=7$ is a solution.