Solve $LaTeX: \displaystyle \log_{10}(x + 117)+\log_{10}(x) = 3$ .

Using logarithmic properties and expanding the argument gives $LaTeX: \displaystyle \log_{10}(x^{2} + 117 x)=3$ . Making both sides an exponent on the base gives $LaTeX: \displaystyle x^{2} + 117 x=10^{3}$ . Expanding and setting equal to zero gives $LaTeX: \displaystyle x^{2} + 117 x - 1000=0$ . Factoring gives $LaTeX: \displaystyle \left(x - 8\right) \left(x + 125\right)=0$ . Solving gives the two possible solutions $LaTeX: \displaystyle x = -125$ and $LaTeX: \displaystyle x = 8$ . The domain of the original is $LaTeX: \displaystyle \left(-117, \infty\right) \bigcap \left(0, \infty\right)=\left(0, \infty\right)$ . Checking if each possible solution is in the domain gives: $LaTeX: \displaystyle x = -125$ is not a solution. $LaTeX: \displaystyle x=8$ is a solution.