# How do you show that tanh^-1(x) = (1/2) ln ((1+x)/(1-x))?

Let $y = {\tanh}^{- 1} x$. Then $x = \tanh y = \frac{{e}^{y} - {e}^{- y}}{{e}^{y} + {e}^{- y}} = \frac{{e}^{2 y} - 1}{{e}^{2 y} + 1}$.. Solving for ${e}^{2 y}$ and then for y, by equating logarithms, we get the logarithmic form.