Insurance company decides to ask for a premium based on: $P = \frac {1} {a} ln(E[e^{aX}])$. Find P for X exponential RV with parameter $\lambda$, and $a = \alpha \lambda$, $0 < \alpha < 1$. (Problem from Ross, Intro to Probability Models 12th)
Read solution as: $E[e^{\alpha \lambda X}] = \int e^{\alpha \lambda x}\lambda e^{-\lambda x} dx = \frac {1} {1 - \alpha}$
Therefore: $P = - \frac {1} {\alpha \lambda} ln(1 - \alpha)$
Even though I read the solution I don't understand how the $\frac {1} {1 - \alpha}$ was obtained, more precisely where did the exponential disappear?
Second question is how come the fraction was inversed in the final solution of P?
To evaluate the integral, you can use the formula $\int_0^\infty e^{-cx} dx = \frac{1}{c}$ (c>0) after a few manipulations to the term.
For your second question, the formula $ln \frac{1}{y} =-ln y$ may answer your question.
If not already done it may help to not only read the solution but also try write down some extra steps to understand the solution. We‘re happy to help if you get stuck somewhere, but please shoe your work.