Show that $e^3 > 20$

If you know your powers of $3$ well, you know $2.7^3=19.683$. Since $e>2.718=2.7\left(1+\frac{2}{300}\right)$,$$e^3>19.683\left(1+\frac{2}{100}\right)=19.683+0.39366>20.$$

Similar to your last proof, I found a positive function whose integral is $e^3-20$.

For $$f(x)=\frac{1}{186}(x-1)^2(x-2)^4e^x\ge0$$

we have $$\int_{0}^{3}f(x)dx=e^3-20$$

$$1+3+\frac92+\frac92+\frac{27}8+\frac{81}{40}+\frac{81}{80}+\frac{243}{560}+\frac{729}{4480}\\ 13+3.375+2.025+1.025+0.433928\cdots+0.162723\cdots=20.021651$$

isn't so difficult. Only the last two term require a "true" division.

An extended comment.

Not really a proof, but an interesting consequence:

$$\log 20=4 \log 2+\log \left(1+\frac{1}{4}\right)<3$$

$$\log 2< \frac34 -\frac14 \log \left(1+\frac{1}{4}\right) $$

$$\log 2< \frac34 -\frac14 \left(\frac{1}{4}-\frac{1}{32}\right) $$

$$\log 2< \frac34 -\frac1{18} $$

The error here is approximately $0.0013$.

That said, there's a lot of inequalities for logarithms, especially for $\log 2$ already known. This can be used to prove the OP.