Need help with an inverse function

Question

$$g(x) = \frac{100}{1+2^{-x}}$$

Ok, i have this expression and my task is to find the inverse. My answer to that is -ln2((100-x)/x). Which is wrong when i test it. Can someone help me with this?

Answer 1

$$g(x) = \frac{100}{1+2^{-x}}$$

$$\frac{100}{g(x)} = 1+2^{-x}$$

$$\frac{100}{g(x)} -1 = 2^{-x}$$

$$\ln\left(\frac{100}{g(x)} -1\right) / \ln(2) = -x$$

$$-\ln\left(\frac{100}{g(x)} -1\right) / \ln(2) = x$$

Answer 2

$g(x)=\frac{100}{1+2^{-x}}\implies$

$g(x)\cdot(1+2^{-x})=100\implies$

$1+2^{-x}=\frac{100}{g(x)}\implies$

$2^{-x}=\frac{100}{g(x)}-1\implies$

$-x=\log_2(\frac{100}{g(x)}-1)\implies$

$x=-\log_2(\frac{100}{g(x)}-1)$

Answer 3

I can let $y=\frac{100}{1+2^{-x}}$.

$$ 1+2^{-x} = \frac{100}{y}$$ $$ 2^{-x} = \frac{100}{y}-1$$ $$ 2^{-x} = \frac{100-y}{y}$$ $$2^x = \frac{y}{100-y}$$ $$x = \log_2 \left ( \frac{y}{100-y}\right)$$