How determine if function f(x)=ln(1-x/1+x) is odd or even?

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You should use the following rules that help you to determine if a function is even or odd, such that:

- `f(-x) = f(x)` (even function)

- `f(-x) = -f(x)` (odd function)

Hence, you should replace -x for x in equation of the function to test its parity, such that:

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

You should notice that you mayreplace `1/((1-x))/(1+x))` for `(1+x))/(1-x)` , such that:

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

Using the logarithmic identities, you need to convert the logarithm of quotient into a difference of logarithms, such that:

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

Since `ln 1 = 0` yields:

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

You should replace f(x) for ln `((1-x)/(1+x))` such that:

`f(-x) = -f(x)`

Since replacing -x for x in equation of the function yields the negative of the function, hence, the function `f(x) = ln ((1-x)/(1+x))` is odd.

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