How can I do step-by-step calculations for the three equations to be equivalent to each other?

Question

I am working on logistic regression, and have faced with the three equations below. I am told they are all equivalent. Why, how can I do the step by step calculations for them to be equivalent?

1. $\dfrac{\Phi}{1-\Phi} = e^{(\beta_0 + \beta_1X_1 + \beta_2X_2 + \dots + \beta_kX_k)}$
2. $\Phi = \dfrac{e^{(\beta_0 + \beta_1X_1 + \beta_2X_2 + \dots + \beta_kX_k)}}{1 + e^{(\beta_0 + \beta_1X_1 + \beta_2X_2 + \dots + \beta_kX_k)}}$
3. $\Phi = \dfrac{1}{1 + e^{-(\beta_0 + \beta_1X_1 + \beta_2X_2 + \dots + \beta_kX_k)}}$

Answer 1

Let $y=e^{(\beta_0 + \beta_1X_2 + \beta_2X_2 + \cdots + \beta_kX_k)}$, then \begin{align} \frac{\Phi}{1-\Phi}&=y\tag1\\ \Phi&=y(1-\Phi)\\ \Phi&=y-y\,\Phi\\ \Phi+y\,\Phi&=y\\ \Phi(1+y)&=y\\ \Phi&=\frac{y}{1+y}.\tag2 \end{align} Multiplying the numerator and denominator of RHS in $(2)$ by $\dfrac1y$ yields \begin{align} \Phi&=\frac{y}{1+y}\cdot\frac{\frac1y}{\frac1y}\\ &=\frac1{\frac1y+1}\\ &=\frac1{y^{-1}+1}.\tag3 \end{align}

Answer 2

Let y = $e^{\sum_{i=0}^k \beta_i X_i}$ with $X_0 \equiv 1$. Equation $1$ says $$ \frac{\Phi}{1 - \Phi} = y. $$ Thus $$ \frac{y}{1+y} = \frac{\Phi/(1-\Phi)}{1 + \Phi/(1-\Phi)} = \frac{\Phi}{(1-\Phi) + \Phi} = \Phi, $$ verifying equation $2$, and equation $3$ is true since $$ \frac{1}{1 + 1/y} = \frac{y}{y+1} $$ which, from above, is again equal to $\Phi$.

Answer 3

Start by rewriting it as $$ \frac{\Phi}{1 - \Phi} = e^B $$ $$ \Phi = \frac{e^B}{1 + e^B} $$ $$ \Phi = \frac{1}{1 + e^{-B}} $$ and then the substitution is simpler to write, even though it is the same thing--just multiply by $e^B$ in the bottom one, for example.

It is always a good idea to simplify notation for any algebraic expression whenever possible. This makes the problem look less intimidating.