Log-transformation of $E(Y|X) = Y_0.exp(\beta_2X)$

Question

Could you explain me why

$E(Y|X) = Y_0.exp(\beta_2X)$

can be transformed into

$E(log(Y)|X) = log(Y_0)+\beta_2X$

?

More specifically, is that equal to $log(E(Y|X))$ and, if so, what's the rule allowing you to go from $log(E(Y|X))$ to $E(log(Y)|X)$?

The right part of the equation is understood.

Thank you!

Answer 1

This is not true. In fact $\log (EY|X) \geq E(\log Y|X)$ and equality holds only in the trivial case $P(Y=EY|X)=1$. [ These are consequences of Jensen's inequality and the fact that logarithm is strictly concave on $(0,\infty)$].