Solve $LaTeX: \displaystyle \log_{ 2 }(x + 4) + \log_{ 2 }(x + 6) = 3$

Using the product rule for logarithms gives $LaTeX: \displaystyle \log_{ 2 }(\left(x + 4\right) \left(x + 6\right))$ and rewriting in exponential form gives $LaTeX: \displaystyle \left(x + 4\right) \left(x + 6\right) = 8$ expanding and setting the equation equal to zero gives $LaTeX: \displaystyle x^{2} + 10 x + 16 = 0$ . Factoring gives $LaTeX: \displaystyle \left(x + 2\right) \left(x + 8\right)=0$ . This gives two possible solutions $LaTeX: \displaystyle x=-8$ or $LaTeX: \displaystyle x=-2$ . $LaTeX: \displaystyle x=-8$ is an extraneous solution. The only soution is $LaTeX: \displaystyle x=-2$ .