Question

How do you find the inverse of 5/e^x+1?

Answer 1

`y=-ln((x-1)/5)`

Explanation:

Remember that `f(x)` is the inverse of `g(x)` if both `f(g(x))` and `g(f(x))` both equal `x`.

From this, we know that when `5/e^x+1=f(x)`, then `5/e^g(x)+1=x` where `g(x)` is the inverse function of `f(x)`.
For simplicity, we will use `y` in place of `g(x)`.

We now isolate `y`.

`5/e^y+1=x`
`5/e^y=x-1`
`5xxe^-y=x-1` We use the fact that `z/x^y=zxx x^-y`
`e^-y=(x-1)/5` We use the natural logarithm on both sides.
`Log _e(e^-y)=Log_e((x-1)/5)` which is really
`ln(e^-y)=ln((x-1)/5)` Remember that `log_x x^y=y`
`-y=ln((x-1)/5)`
`y=-ln((x-1)/5)` That is your answer!