已知$n\in\mathbb N^*$，$x=\left(1+\dfrac 1n\right)^n$，$y=\left(1+\dfrac 1n\right)^{n+1}$，则( )
A．$x^y>y^x$
B．$x^y=y^x$
C．$x^y<y^x$
D．$x^y$和$y^x$的大小关系跟$n$的取值有关
正确答案是B．
分析与解 根据题意，有\[y\ln x=\left(1+\dfrac 1n\right)^{n+1}\cdot n\ln \left(1+\dfrac 1n\right)=\left(1+\dfrac{1}n\right)^n\cdot ( n+1)\ln\left(1+\dfrac 1n\right)=x\ln y,\]所以选项B正确．
也可以对要比较的两个数取对数之后算比值，有$$\dfrac {\ln{x^y}}{\ln{y^x}}=\dfrac {y\ln x}{x\ln y}=\dfrac {\left(1+\dfrac 1n\right)^{n+1}\cdot n\ln\left(1+\dfrac 1n\right)}{\left(1+\dfrac 1n\right)^n\cdot(n+1)\ln\left(1+\dfrac 1n\right)}=1.$$