# Verification of Logistic CDF with Universality of the Uniform

I'm studying Continuous Random Variable in the topic Universality of the Uniform.

From the book "Introduction to probability" by Joseph K. Blitzstein and Jessica Hwang page 207-208 the writer shows a calculation to verify that $F^{-1}(U) = F(x)$ for logistic CDF: now let U ~ Uniform(0,1)

The Logistic CDF is:

$$F(x) = \frac{e^x}{1+e^x} , x \in R$$

First we find its inverse:

$$F^{-1}(u) = log(\frac{u}{1-u})$$

Then we plug in U for u:

$$F^{-1}(U) = log(\frac{U}{1-U})$$

Now lets do calculation step by step to verify that $F^{-1}(U) = F(x)$

$$F^{-1}(U) = log(\frac{U}{1-U})$$ $$P(log(\frac{U}{1-U}) \le x) = P(\frac{U}{1-U} \le e^x)$$ $$= P(U \le e^x(1-U))$$ $$= P(U \le \frac{e^x}{1+e^x})$$ $$= \frac{e^x}{1+e^x}$$

I understand most of this except in the line that says $P(U \le e^x(1-U)) = P(U \le \frac{e^x}{1+e^x})$ How can the $e^x(1-U) = \frac{e^x}{1+e^x}$ can someone please explain to me step by step with easy detail explanation. I'm newbie to Probability.

$$U \leq exp(x) (1-U)$$
$$U \leq \exp(x) - \exp(x)U$$
$$(1+\exp(x))U \leq \exp(x)$$
$$U \leq \frac{\exp(x)}{1+\exp(x)}$$