Let $X$ be an exponential random variable with parameter $\lambda = 4$, and let $Y$ be the random variable defined by $Y=8e^X$. Compute the probability density function of $Y$:
$f_Y(t) = $
Hint: first find the cdf of $Y$ and use the cdf of $X$ to help with the computation. $$P(Y \le y)=?$$
$$P(Y \le y) = P(8e^X \le y) = P(X \le \log(y/8))$$
Once you have the cdf $P(Y \le y)$, take the derivative with respect to $y$ to get the pdf.