<s>[INST] Compare $e^2$ and $7$ without using calculator Which is bigger? $e^2$ or $7$? Any tricks?
Don't know quite how to approach those kind of things. 
 [/INST] Here's an alternative. Try to compare $2$ and $\ln(7)$. $\ln(x)$ has a more natural geometric interpretation as area under $y=\frac{1}{x}$, compared to $e^x$, so some geometry might avail us. 
Since $\ln(7)=\int_1^7\frac{1}{x}\,dx$, we consider approximating the integral with a trapezoid rule style Riemann sum. The trapezoids closer to $1$ will do a worse job since concavity is higher there, so we want these trapezoids to have a narrower base. Using a linear progression of base width would mean we need to cut up $[1,7]$ into $T_n$ equal sized pieces, where $T_n$ is a triangular number. Using $T_n=15$ (and therefore $5$ trapezoids with bases $\frac{2}{5}$, $\frac{4}{5}$, $\frac{6}{5}$, $\frac{8}{5}$, $\frac{10}{5}$) gives us:

And we can see/calculate that $$\ln(7)<\frac{1}{2}\left(\frac{2}{5}\left(1+\frac{5}{7}\right)+\frac{4}{5}\left(\frac{5}{7}+\frac{5}{11}\right)+\frac{6}{5}\left(\frac{5}{11}+\frac{5}{17}\right)+\frac{8}{5}\left(\frac{5}{17}+\frac{1}{5}\right)+\frac{10}{5}\left(\frac{1}{5}+\frac{1}{7}\right)\right)$$
$$\ln(7)<\frac{65376}{32725}<2$$
So $7<e^2$.
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