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 $\displaystyle X \sim \mbox{exp}(\theta)$ (exponential distribution) where the distribution function $\displaystyle F$ is given by $\displaystyle F_{X}(x) = 1-e^{-\theta x}$ (for $\displaystyle x \geq 0$). We can compute the distribution function of the random variable $\displaystyle Y=e^X$ as follows: $\displaystyle F_{Y}(x) = P(Y \leq x) = P(e^X \leq x) = P(X \leq \ln(x)) = F_{X}[\ln(x)] $ therefore

$\displaystyle F_{Y}(x) = 1-e^{-\theta \ln(x)}$. We can find the pdf of $\displaystyle Y$ (let's call it $\displaystyle g(x)$) by calculating the first derivative of the distribution function, hence $\displaystyle g(x) = \frac{dF_{Y}(x)}{dx} = \frac{d}{dx}\left[1-e^{-\theta \ln(x)}\right] = \left(\frac{\theta}{x}\right)e^{\theta \ln(x)}$