What range of values of lambda does the mean of $Y$ converge? What is the mean in that case? $X$ is an exponential RV with parameter lambda and $Y = e^x$. So, I found the density of $Y$ to be $\lambda y^{-\lambda}e^{-y}$. Then to find the range of $\lambda$ where the mean converges, do we just take the integral of the mean from zero to infinite and see where it would converge? Is that the correct process so far?
 A: Your density is not correct.  If $Y$ is a one-to-one function $g$ of another random variable $X$--that is, if $Y = g(X)$ for some invertible function $g$, then $$f_Y(y) = f_X(g^{-1}(y)) \left| \frac{dg^{-1}}{dy} \right|.$$  In your case, $$f_X(x) = \lambda e^{-\lambda x}, \quad x > 0,$$ and $Y = g(X) = e^X$, hence $X = g^{-1}(Y) = \log Y.$  Given this, what is your density of $Y$?
Now consider the support of $Y$:  namely, if $X > 0$, then what are the possible values of $Y$?  Call the support $\Omega$.
Next, consider the expectation of $Y$:  you could integrate it directly via the formula $$\operatorname{E}[Y] = \int_{y \in \Omega} y f_Y(y) \, dy.$$  For what values of the parameter $\lambda$ would this interval converge?
Alternatively, you may do the computation directly via the moment generating function of the random variable $X$:  recall that $$M_X(t) = \operatorname{E}[e^{tX}],$$ for all $t$ for which such an expectation is defined.  But since $Y = e^X$, it is easy to see that $$\operatorname{E}[Y] = M_X(1).$$  Now recall that the MGF of an exponential distribution with rate parameter $\lambda$ is $$M_X(t) = \frac{\lambda}{\lambda - t}, \quad t < \lambda.$$  So, under what conditions of $\lambda$ would this MGF exist at $t = 1$?
A: By the Law of the Unconscious Statistician, we have
$$E(e^X)=\int_0^\infty e^x \lambda e^{-\lambda x}\,dx.$$
Since $e^x e^{-\lambda x}=e^{-(\lambda-1)x}$, the integral converges if and only if $\lambda\gt 1$. 
Calculation of the integral, when it converges, is straightforward. We get $\frac{\lambda}{\lambda-1}$.
