As you said, notice that 17.4 = 16 + 1.4, and the only hard bit is to calculate 1.4 / 4
To divide 1.4 by 4, use the fact that dividing by 4 is the same as dividing by 2 twice. If you divide by 1.4 by 2 once, you get 0.7. If you divide by 2 again, you get 0.35, so the answer is 4.35
It might help, when dealing with decimals, to multiply them by a power of 10 in your head before doing the division. For example, when calculating 1.4 / 2 I mentally convert that to 14 / 2 (which is 7) and then divide by 10 again to get 0.7
Now to do 0.7 / 2, I multiply by 10 to get 7 / 2 (which is 3.5) and then divide by 10 to get 0.35
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
You can use the rough approximation of $10^3 \approx 2^{10}$ for 1sf computations.
$\log_2(10) = \frac{1}{3} \log_2(1000) \approx \frac{1}{3} \log_2(1024) = \frac{10}{3}$.
If you want more precision, remember that $\ln(1+x) \approx x - \frac{1}{2} x^2$ as $x \to 0$, so taking just the first-order term we get $\log_2(1.024) = \frac{\ln(1.024)}{\ln(2)} \approx \frac{0.024}{\ln(2)}$. Now you can either use $\ln(2) \approx .693$ for 3sf precision, or you can use a rational approximation $\ln(2) \approx \frac{9}{13}$, which gives $\log_2(1.024) \approx \frac{104}{3000}$, and hence $\log_2(10) \approx \frac{10}{3} - \frac{104}{9000}$, which is good to 4 significant figures!
And from that we get $\log_2(10^{20}) \approx \frac{200}{3} - \frac{208}{900}$.