Functions of one random variable question

I have a problem as follows:

Let X have the pdf f(x)= θ * e^{( - θ x)}, 0 < x < infinity

find the pdf of Y = e^{x}

I've gone about this the way I normally do for these problems.

I have

G(y) = P(X < ln y) = integral of θ * e^{( - θ x)} dx from 0 to ln y = 1 - y^{-}^{θ}^{ }

So then I need g(y) = G'(y) of Y? which would be

θ * e^{( - θ (ln y)}??

or am I missing something super-duper obvious?

Thanks

Re: Functions of one random variable question

Note that the random variable (exponential distribution) where the distribution function is given by (for ). We can compute the distribution function of the random variable as follows: therefore

. We can find the pdf of (let's call it ) by calculating the first derivative of the distribution function, hence