It depends on what you have memorized. For $\log_2 10$, many people know that $\log_{10} 2\approx 0.30103$, so $\log_2 10 = \frac 1{0.30103}=\frac {10}{3}-\frac 13\% \approx 3.32$ where the last $\frac 13\%$ is optional and accounts for the difference between $0.3$ and $0.301$. The correct value is about $0.3322$
Many base $2$ logs are easier because you might know some powers of $2$
For $\log_3 10$ we know $3^2=9$ so it should be a little more than $2$. I think my next step would be to say $3^{2.5}=9*1.732=17.32-1.73=15.59$ As $10$ is about $\frac 17$ of the way from $9$ to $16$ I would add $\frac {0.5}7=0.07$ and say $\log_3 10 \approx 2.07$ The correct value is about $2.10$ Not too bad for a long linear interpolation.
I did play fair and not check the correct result until afterward. These are fairly hard as estimates go. Multiplication and division are much easier as one is more prone to have useful facts memorized. $\ln 10=2.30$ and $\ln 2=0.693$ can be useful, too.
If you want to stay within 10%, the following piecewise linear function satisfies $$.9\tan\theta \le y \le 1.1\tan\theta$$ for $0\le\theta\le60$ degrees:
$$y={\theta\over60}\text{ for }0\le\theta\le20$$
$$y={2\theta-15\over75}\text{ for }20\le\theta\le45$$
$$y={\theta-20\over25}\text{ for }45\le\theta\le60$$
It might help to rewrite them as
$$y={5\theta\over300}\text{ for }0\le\theta\le20$$
$$y={8\theta-60\over300}\text{ for }20\le\theta\le45$$
$$y={4\theta-80\over100}\text{ for }45\le\theta\le60$$
so that you really don't have to divide by anything other than $3$. The line segment approximations lie above $\tan\theta$ from $\theta\approx25$ to $\theta=45$ and below it elsewhere, so you should round down and up accordingly when doing the mental arithmetic.It's obviously possible to extend this for angles greater than $60$ degrees, but whether (or how far) you can do so with formulas that use only "simple" multiplications and divisions is unclear.
A word of explanation: What I tried to do here was take seriously the OP's request for estimates you can calculate in your head. The ability to do mental arithmetic, of course, varies from person to person, so I used myself as a gauge. As for where the formulas came from, my starting point was the observation that the conversion factor between degrees and radians, $180/\pi$, is approximately $60$, so the estimate $\tan\theta\approx\theta/60$ should be OK for a while. A little trial and error showed it's good up to $\theta=20$ degrees (since $.9\tan20\approx.328$). It was easy to see that connecting $(0,0)$ to $(20,1/3)$ and $(20,1/3)$ to $(45,1)$ with straight lines would stay within the prescribed bounds. Finally, noting that $.9\tan60\approx1.55$, I saw that the line connecting $(45,1)$ to $(60,1.6)$ would have a nice slope and stay within the prescribed bounds as well.
Best Answer
Using calculus, specifically Taylor approximation, centered at 8 (which is the closest value to 9.1 whose cube root we know), we get $$(8+x)^{1/3}=2+\frac{1}{12}x-\frac{1}{288}x^2+\frac{5}{20736}x^3\cdots$$ Hence, in your example, we have $x=1.1$, or $$(8+1.1)^{1/3}=2+\frac{1}{12}1.1-\frac{1}{288}1.1^2+\frac{5}{20736}1.1^3\cdots$$ You can compute as many terms as you need. One term givs you 2. Two terms gives you $2.091\overline{6}$. Three terms gives you $2.0874652\overline{7}$.