An interview question: $2.1^{3.1}$ vs $3.1^{2.1}$,$ 2.1^{4.1}$ vs $4.1^{2.1}$, which is larger? While Mathematica told me that $2.1^{3.1} - 3.1^{2.1} = -0.786932$ and $2.1^{4.1} - 4.1^{2.1} = 1.58855$, I wonder how to compare them quickly, by hand.
I see $2^3 < 3^2$, so perhaps we have $2.1^{3.1} < 3.1^{2.1}$, but I'm not sure.
As for $2^4 = 4^2$, I guess $2.1^{4.1}$ is larger because the exponent is bigger, but I'm not too sure.

Let's consider
$$(a+\epsilon)^{b+\epsilon} = e^{(b+\epsilon) \log{(a+\epsilon)}}$$
$$\begin{align}(b+\epsilon) \log{(a+\epsilon)} &= (b+\epsilon) \left [\log{a} + \log{\left (1+\frac{\epsilon}{a} \right )} \right ] \\ &= (b+\epsilon) \left [\log{a} +\frac{\epsilon}{a} - \frac12 \frac{\epsilon^2}{a^2} +O\left (\epsilon^3 \right ) \right ]\\ &= b \log{a} + \left (\frac{b}{a}+\log{a} \right )\epsilon + \left (\frac1{a}-\frac{b}{2 a^2} \right )\epsilon^2 + O\left (\epsilon^3 \right )\end{align}$$
Exponentiating, we get that
$$(a+\epsilon)^{b+\epsilon} = a^b \left [1+ \left (\frac{b}{a}+\log{a} \right )\epsilon + \left (\frac1{a}-\frac{b}{2 a^2}+ \frac12 \left ( \frac{b}{a}+\log{a} \right )^2 \right ) \epsilon^2 \right ]+ O\left (\epsilon^3 \right )$$
For the purposes of the interview, we only need look at the $O(\epsilon)$ term.  Thus, when $a=2, b=3$ and $\epsilon=0.1$, we get that $2.1^{3.1} \approx 8 + 8 (1.5+\log{2}) (0.1) \approx 9.8$.  In contrast, when $a=3, b=2$ and $\epsilon=0.1$, we get that $3.1^{2.1} \approx 9 + 9 (0.7+\log{3}) (0.1) \approx 10.6$.  The difference is sufficiently greater than $a^b \epsilon^2$ that we don't need to consider the $O(\epsilon^2)$ term.  Thus, we can say that $3.2^{2.1} \gt 2.1^{3.1}$.
The OP should be able to continue the analysis for the other case, which seems more interesting.
