Finding p.d.f. of a function of random variable which is exponentially distributed

Question

Given random random variable $X \sim \mathrm{Exp}(\lambda)$, find p.d.f of $Y=\ln(X)$

Approach

I decided to go the cumulative distribution route. I take integral of p.d.f. of $X$ with $y$:

$$F(y) = \int_{0}^{y} \lambda \exp(-\lambda \cdot x) = -\exp(-\lambda \cdot \ln(x)) + 1=1 - y^{-\lambda}.$$

Then, take first derivative

$$f_Y(y)=F'(y)=\lambda y^{-\lambda}.$$

But the answer is completely different:

$$f_Y(y)=\lambda e^{y-\lambda e^{t}}.$$

(I changed variable letters of the answer to match letters in my solution, hopefully I didn't mess up anything)

Question

Where did I make a mistake? Did I do something completely wrong?

I've checked the integraion with integral calculator, but it seems to be correct.

Answer 1

You've miscalculated $F$ to start with. Here's a hint: $$F(y) = P[Y \leq y] = P[\ln(X) \leq y] = P[X \leq e^y]$$