8. 已知 $a,b$ 满足 $ae^{a}=e^{3}$，$b\ln b=e^{4}+b$，则（ ）
A. $\dfrac{5}{e}<a<\mathrm{e}$ B. $ab>e^{4}$ C. $b<\mathrm{e}^{2}a$ D. $\dfrac{2}{5}\mathrm{e}^{4}<b<\dfrac{\mathrm{e}^{4}}{2}$
解：$\because ae^{a}=e^{3}$，$b\ln b=e^{4}+b$
$\therefore a>1,b>1$
$\therefore e^{b\ln a}e^{a}=e^{3}$，$b(b\ln b-1)=e^{4}$
$\therefore e^{\ln a+a}=e^{3}$，$\dfrac{b}{e}\ln\dfrac{b}{e}\cdot e^{\frac{\ln b}{e}}=e^{3}$
$\therefore \ln a+a=3$，$ae^{a}=\ln\dfrac{b}{e}\cdot e^{\frac{\ln b}{e}}$
设 $f(x)=xe^{x}$，则 $f'(x)=(x+1)e^{x}$
$\therefore x>0$ 时 $f(x)$ 单增
$\therefore a=\ln\dfrac{b}{e}$
$\therefore b=e^{a+1}$
设 $g(x)=\ln x+x$，显然 $g(x)$ 单增
$\because g(2)=\ln 2+2<3$，$g(2.5)=\ln 2.5+2.5>\ln\sqrt{e}+2.5=3$
$\because g(a)=3$
$\therefore 2<a<2.5$，故 A 错误
设 $t(x)=xe^{x+1}$，由上可知 $x>0$ 时 $t(x)$ 单增
$\because ab=ae^{a+1}$
$\therefore ab<2.5e^{3.5}=\dfrac{5}{2\sqrt{e}}e^{4}<e^{4}$。故 B 错误
$\because \dfrac{b}{a}=e^{a}$
$\therefore \dfrac{b}{a}>e^{2}$
$\therefore b>ae^{2}$。故 C 错误
$\because 3+\ln\dfrac{2}{5}-2=1+\ln\dfrac{2}{5}=\ln\dfrac{2e}{5}>0$
$3+\ln\dfrac{2}{5}-2.5=\dfrac{1}{2}+\ln\dfrac{2}{5}=\ln\dfrac{2\sqrt{e}}{5}<0$
$\therefore 3+\ln\dfrac{2}{5}<a<2.5$
$\therefore e^{3+\ln\frac{2}{5}}<b<e^{3.5}$
$\therefore \dfrac{2}{5}e^{4}<b<\dfrac{1}{\sqrt{e}}e^{4}<\dfrac{1}{2}e^{4}$，故 D 正确