Solve $LaTeX: \displaystyle \log_{ 15 }(x + 20) + \log_{ 15 }(x + 3380) = 4$

Using the product rule for logarithms gives $LaTeX: \displaystyle \log_{ 15 }(\left(x + 20\right) \left(x + 3380\right))$ and rewriting in exponential form gives $LaTeX: \displaystyle \left(x + 20\right) \left(x + 3380\right) = 50625$ expanding and setting the equation equal to zero gives $LaTeX: \displaystyle x^{2} + 3400 x + 16975 = 0$ . Factoring gives $LaTeX: \displaystyle \left(x + 5\right) \left(x + 3395\right)=0$ . This gives two possible solutions $LaTeX: \displaystyle x=-3395$ or $LaTeX: \displaystyle x=-5$ . $LaTeX: \displaystyle x=-3395$ is an extraneous solution. The only soution is $LaTeX: \displaystyle x=-5$ .