Converting $\tanh^{-1}{x}$ to an expression involving the natural logarithm I know how to convert $\sinh^{-1}{x}$ and $\cosh^{-1}{x}$ to $\ln{|x+\sqrt{x^2 \pm 1}|}$, but for some reason I am struggling to do the same for the following statement: 
$$\tanh^{-1}{\frac{x}{2}}$$
Can someone please show me how to convert it to a $\ln$ form? thanks!

Set $ y = \tanh^{-1} t $ and take $\tanh$ to take both sides so we have $$ \tanh y = t .$$
Now convert the $\tanh$ term into it's definition in terms of exponentials: $$ t = \frac{ e^{y} - e^{-y} }{e^y + e^{-y} } = \frac{ e^{2y} -1 }{e^{2y} +1 }.$$
Remember we want to solve for $y.$ Firstly solve for $u= e^{2y} $ first. Rearranging $ t= \frac{u-1}{u+1} $ is simple, and hopefully you can do the rest. 
