Solve LaTeX:  \displaystyle \log_{15}(x + 3122)+\log_{15}(x + 240) = 5 .

Using logarithmic properties and expanding the argument gives LaTeX:  \displaystyle \log_{15}(x^{2} + 3362 x + 749280)=5 . Making both sides an exponent on the base gives LaTeX:  \displaystyle x^{2} + 3362 x + 749280=15^{5} . Expanding and setting equal to zero gives LaTeX:  \displaystyle x^{2} + 3362 x - 10095=0 . Factoring gives LaTeX:  \displaystyle \left(x - 3\right) \left(x + 3365\right)=0 . Solving gives the two possible solutions LaTeX:  \displaystyle x = -3365 and LaTeX:  \displaystyle x = 3 . The domain of the original is LaTeX:  \displaystyle \left(-3122, \infty\right) \bigcap \left(-240, \infty\right)=\left(-240, \infty\right) . Checking if each possible solution is in the domain gives: LaTeX:  \displaystyle x = -3365 is not a solution. LaTeX:  \displaystyle x=3 is a solution.