Comparing huge numbers!

$\begin{aligned} \text{Consider } \frac{99^{e^{101}}}{101^{e^{99}}} & = \frac{\ln (99^{e^{101}})}{\ln (101^{e^{99}})} =\frac{e^{101} \ln 99}{e^{99} \ln 101} = \frac{e^{99}e^2 \ln 99}{e^{99} \ln 101} = \frac{e^{2} \ln 99}{\ln 101} < \frac{e^{2} \ln 99}{\ln 99} = e^{2} > 1 \\ \Rightarrow 99^{e^{101}} & \boxed{>} \space 101^{e^{99}} \end{aligned}$

Consider $f(x) = e^{-x}\ln{x}$ . This function is definitely decreasing in $[e,\infty ]$ . So $f(99) > f(101)$

If two numbers are subtracted and the difference is positive, $a - b > 0$ , then the smaller number was subtracted from the larger, $a > b$ . Taking the natural log of both yields $e^{101} \times ln(99)$ and $e^{99} \times ln(101)$ using an equail sign as a placeholder. Dividing $e^{99}$ from both gives us $e^{2} \times ln(99)$ and $ln(101)$ By subtracting, we get $e^{2} ln(99) - ln(101)$ which is a positive number. Therefore, $e^{101} \times ln(99) > e^{99} \times ln(101)$ which gives us $\boxed{99^{e^{101}} > 101^{e^{99}}}$ .