Log's power is log itself (1)

Both $ln(2.1)$ and $ln(2.2)$ are positive and hence we raise both sides of the inequality to the power $\frac{1}{ln(2.1)ln(2.2)}$ without affecting the sign of the given inequality.

$\displaystyle \Rightarrow \big( ln(2.1)\big) ^{\frac{1}{ln(2.1)}} > \big( ln(2.2)\big) ^{\frac{1}{ln(2.2)}}$

Now consider the function $f(x) = x^{\frac{1}{x}}$

$f'(x) = x^{\frac{1}{x}} \big( \frac{1-lnx}{x^2} \big)$

$f'(x) > 0$ in $x \in (0,e)$ .

Since $f$ is increasing in $(0,e)$ :

$0 < ln(2.1) < ln(2.2) < e$

$\displaystyle \Rightarrow f(0) < f(ln(2.1)) < f(ln(2.2)) < f(e)$

$\displaystyle \Rightarrow \big( ln(2.1)\big) ^{\frac{1}{ln(2.1)}} < \big( ln(2.2)\big)^{\frac{1}{ln(2.2)}}$

Hence, the given inequality is false.