[tex]Y=2e^X\iff X=\ln\dfrac Y2[/tex]
Note that [tex]x\in[0,\infty)[/tex], so that [tex]y\in[2,\infty)[/tex].
[tex]F_Y(y)=\mathbb P(Y\le y)=\mathbb P(2e^X\le y)=\mathbb P\left(X\le\ln\dfrac y2\right)=F_X\left(\ln\dfrac y2\right)[/tex]
[tex]F_X(x)=1-e^{-4x}[/tex]
[tex]\implies F_Y(y)=1-e^{-4\ln\frac y2}=1-\dfrac{16}{y^4}[/tex]
[tex]\implies f_Y(y)=\dfrac{\mathrm dF_Y(y)}{\mathrm dy}=\dfrac{64}{y^5}[/tex]
