I would much appreciate if you help me out with this problem
Let $X \sim Unif(0,1)$
Find the density of $Y = -\lambda^{-1} \log(1-X)$ with $\lambda > 0$
Then calculate $P(Y>t+s|Y>t)$ for $t,s >0$
So far i´ve got
$F_Y (y) = P[Y \le y]$ $=$ $P[1/y^{\lambda} \le log(1-x)]$ $=$
$P[X \le 1-10^{1/y^{\lambda}}]$ this is $F_X(1-10^{1/y^\lambda})$
Im stuck there. Any suggestions or help?
The approach through the cumulative distribution function is reasonable, but there are troubles with the algebra. I am interpreting $\log$ as the natural logarithm. Minor modification will take care of things if we interpret $\log$ as logarithm to the base $10$.
We have for suitable $y$ $$\Pr(Y\le y)=\Pr\left(-\frac{\log(1-X)}{\lambda}\le y\right)=\Pr(\log(1-X)\ge -\lambda y).\tag{1}$$ The right-hand side of (1) is equal to $$\Pr(1-X\ge e^{-\lambda y}),$$ which is $$\Pr(X\le 1-e^{-\lambda y}).$$ For $y\ge 0$, this is $1-e^{-\lambda y}$ (for $y\lt 0$ it is $0$).
Thus the cdf of $Y$ is $1-e^{-\lambda y}$ if $y\ge 0$, and $0$ otherwise. Differentiate to find the density. But from the cdf we recognize the exponential distribution with parameter $\lambda$.