How can I solve the equation...

Question

$$\log\left(\frac{\pi_i}{1-\pi_i}\right)=\sum_{k=0}^K x_{ik}\beta_k\qquad i=1,2,\dots,N$$

How can I make the equation above the one below by taking "$e$" to both sides.

Note that after taking $e$ to both sides of Eq. 1, $$\frac{\pi_i}{1-\pi_i}=e^{\sum_{k=0}^K x_{ik}\beta_k}.$$

Thank you in advance.

Answer 1

$$\log{\frac{\pi_i}{1-\pi_i}}=\sum\limits_{k=0}^Kx_{ik}\beta_k$$ , take the exponential of both sides, $$e^{\log{\frac{\pi_i}{1-\pi_i}}}=e^{\sum\limits_{k=0}^Kx_{ik}\beta_k}$$ , use the rule $e^{\log{x}}=x$, $$\frac{\pi_i}{1-\pi_i}=e^{\sum\limits_{k=0}^Kx_{ik}\beta_k}$$

Answer 2

The answer comes from the equality $e^{\log A}=A$