Solve $LaTeX: \displaystyle \log_{ 15 }(x + 20) + \log_{ 15 }(x + 230) = 3$

Using the product rule for logarithms gives $LaTeX: \displaystyle \log_{ 15 }(\left(x + 20\right) \left(x + 230\right))$ and rewriting in exponential form gives $LaTeX: \displaystyle \left(x + 20\right) \left(x + 230\right) = 3375$ expanding and setting the equation equal to zero gives $LaTeX: \displaystyle x^{2} + 250 x + 1225 = 0$ . Factoring gives $LaTeX: \displaystyle \left(x + 5\right) \left(x + 245\right)=0$ . This gives two possible solutions $LaTeX: \displaystyle x=-245$ or $LaTeX: \displaystyle x=-5$ . $LaTeX: \displaystyle x=-245$ is an extraneous solution. The only soution is $LaTeX: \displaystyle x=-5$ .