# How do you solve  2 ln (x + 3) = 0 and find any extraneous solutions?

This equation means that $x + 3 = 1$ or more precisely $x + 3 = {e}^{2 i \frac{\pi}{n}}$. For x real $x + 3 = \pm 1$ thus $x = - 4 \mathmr{and} x = - 2$