Expected value of $X$ if $\ln(X)\sim N(\mu, \sigma)$ I am very new to these statistical concepts and getting confused really quickly while reading online.
I am supposed to find the expectation of $\ln(X)\sim N(\mu, \sigma)$ by using the MGF.
So far I've got
$M_{\ln(X)}(t)=E[e^{t\ln(X)}]$ (By defintion) $=E[X^t]$
but I am not sure how to go from there.
I saw someone online simply plugging in $1$ for $t$ and saying that's the expectation but I didn't find anything on why is that and I don't know if it's correct.
Another thing I know is that the expectation is the first derivative of the MGF evaluated at $t=0$. But I am not sure how to use that. Do I simply plug in $\ln(x)$ instead of $x$ in the MGF of $X\sim N(\mu,\sigma)$ then take the first derivative? or?

The moment-generating function of a $N(\mu, \sigma)$ random variable is $e^{\mu t + \frac{\sigma t^2}{2}}$, assuming that $\sigma$ is the variance. If you meant that $\sigma$ is the standard deviation, then $\sigma^2 $ is the variance, and then you need to put $\sigma^2$ in place of $\sigma$ above.
Given the assumed distribution of $\log X$ as you've described, it is known that $M_{\log X} (t) = e^{\mu t + \frac{\sigma t^2}{2}}$.
If you want to find the expectation of $\log X$, you can differentiate the above expression and then plug in $t = 0$.
However, if you wanted to find the expectation of $X$, then you use what you were getting at, namely that $E(X^t) = e^{\mu t + \frac{\sigma t^2}{2}}$, and then you can certainly plug in $t = 1$ to find the answer.
