Show that $Y_i=\frac {e^{\beta_0+\beta_1X_1}}{1+e^{\beta_0+\beta_1X_1}}$ can be linearized by ln $$Y_i=\large \frac {e^{\beta_0+\beta_1X_1}}{1+e^{\beta_0+\beta_1X_1}}$$
can be linearized to $\displaystyle\ln\left(\frac {Y_i}{1-Y_i}\right)=\beta_0+\beta_1X_i$
I did some algebraic manipulation but could not show it. I have problem breaking up the denominator of the RHS.
 A: Hint: Show and use that
$$
1-Y_i=\frac{1}{1+e^{\beta_0+\beta_1X_1}}.
$$
A: Compute Exp[b0+b1 x], which appears in the numerator and the denominator, as a function of Y. That is to say, set Z =  Exp[b0+b1 x]; then Y = Z / (1 + Z). Solve for Z. Replace Z by its definition and take logarithms of both sides. 
A: $$
y = \frac{e^w}{1+e^w}.
$$
Multiplying both the numerator and the denominator by $e^{-w}$ we get
$$
y = \frac{e^w}{1+e^w} = \frac{1}{e^{-w}+1}.
$$
The advantage of this form is that $w$ appears only once, so we need only invert a sequence of functions until $w$ is isolated.  First take reciprocals of both sides:
$$
\frac1y = e^{-w}+1.
$$
Then subtract $1$ from both sides:
$$
\frac1y - 1 = e^{-w}.
$$
Then take logs of both sides:
$$
\log\left(\frac1y - 1\right) = -w.
$$
Then multiply both sides by $-1$:
$$
-\log\left(\frac1y - 1\right) = w.
$$
Then do some routine algebra:
$$
w = -\log\left(\frac1y - 1\right) = -\log\left(\frac{1-y}{y}\right) = \log\left(\frac{y}{1-y}\right).
$$
The function $y\mapsto\log\left(\frac{y}{1-y}\right)$ is called the logit --- the first syllable sounds like "low" and rhymes with "slow", and the "g" is pronounce as in "general", not as in "get":
$$
\operatorname{logit}(y) = \log\left(\frac{y}{1-y}\right).
$$
