# I wonder if you could show me if f(x)=ln1-x/1+x is odd.

*print*Print*list*Cite

### 1 Answer

You need to remember that you may prove that a function is odd if `f(-x) = -f(x)` (hence the graph of the function is symmetric about the origin.

You should substitute -x for x in equation of function such that:

`f(-x) = ln((1-(-x))/(1-x))`

`f(-x) = ln((1+x)/(1-x))`

You may write `((1+x)/(1-x)) as ((1-x)/(1+x))^(-1)` such that:

`ln((1+x)/(1-x)) =ln(((1-x)/(1+x))^(-1))`

You need to use power property such that:

`ln(((1-x)/(1+x))^(-1)) = -ln((1-x)/(1+x))`

Notice that `ln((1-x)/(1+x))` expresses the equation of function f(x), hence`f(-x) = -f(x).`

**Hence, substituting -x for x in equation of function yields f(-x)=-f(x), thus the function has the property of an odd function.**