As in title given that $X$ is uniform in $[0,1]$ and $Y = -\frac{1}{\lambda}\log X$ ($\lambda \gt 0$), to find the density of $Y$ I'm attempting in the following way: finding the cumulative distribution of $Y$ and then deriving to obtain the density of $Y$.
$$P(Y \le t) = P\left(-\frac{1}{\lambda}\log X \le t\right) = P(X \le e^{-\lambda t})$$
By intuition I think that it will be an exponential of parameter $\lambda$ (I know that an $exp(\lambda)$ has cumulative distribution $F(t) = 1-e^{-\lambda t}$.
But when I compute
$$P(X \le e^{-\lambda t}) = \int_0^{e^{-\lambda t}} dx$$
I didn't get what I expected (in fact I get $e^{-\lambda t})$. For sure I'm making mistakes about the integrand function or the integration boundaries.. I probably still have some uncertainties about the uniform distribution. If $X \sim \mathcal{U}[0,1]$ then $F_X(t) = t$ if $t \in [0,1]$,while $f_X(t) = 1$ for $t \in [0,1]$ and $0$ otherwise... right?
in a very simple way,
$$f_Y(y)=\lambda e^{-\lambda y}$$
that is
$$f_Y(y)\sim Exp(\lambda)$$
your error is this
$$P(Y\leq y)=P(X>e^{-\lambda y})=1-F_X(e^{-\lambda y})=1-e^{-\lambda y}$$
In this case the integral is not needed because the density of Y can be derived immediately calculating the derivative of $X=g^{-1}(Y)$:
$$f_Y(y)=\Bigg|\frac{d}{dy}e^{-\lambda y}\bigg|$$