estimate value of $\sqrt[30]{0.05}$ Yesterday I got an exam in which there was a problem and its solution results in $$\sqrt[30]{0.05}$$
I didn't go further calculation. Still I can't.
My lecturer said, even I'm still not sure if he made ironic humor, that if there is a math operation requiring higher than mid-school level math you to do, not do that, let it leave as it is. As a note, my department is not math.
Did he really make a humor?
Is there way/methods to figure out/estimate its result without calculator w.r.t thinking in exam(time limit) and not in exam?
 A: We know that $\sqrt[30]{0.05}$ is a number a little smaller than $1$ because $\sqrt[n]{0.05}$ converges to $1$ for $n$ to infinity. So set $\sqrt[30]{0.05}=1-a$ and then try to estimate $a$. $a$ satisfies the equation $$(1-a)^{30}=\frac{1}{20}$$ Writing out the first few terms gives $$1-30a+\frac{30\cdot 29}{2}a^2 + ...= \frac{1}{20}$$ Note that because $a$ is small, the further coefficients are decreasing quickly. 
Using just $1-30a\sim \frac{1}{20}$ yields $\sqrt[30]{0.05} \sim 0.93$ without calculator. If you use more terms you should get better approximations.
Edit: It turns out this doesn't work quite as well as I thought. While it is true that the later coefficients with higher powers of $a$ are decreasing quickly, the biggest coefficient in the series is at $a^3$. So in order to get something that is actually an approximation of $a$ one would have to compute at least until $a^4$ or $a^5$. This leads to a polynomial which is not really easy to solve by hand. Computing further terms would increase accuracy but I'm not sure whether this is helpful in a no calculator scenario. 
A: You can take it step-by-step:
$$\sqrt[30]{0.05}=\sqrt[5]{\sqrt[3]{\sqrt{\frac5{100}}}}\approx\sqrt[5]{\sqrt[3]{\frac{2.2}{10}}}=\sqrt[5]{\sqrt[3]{\frac{220}{1000}}}\approx \sqrt[5]{\frac{6}{10}}=\sqrt[5]{\frac{60000}{100000}}\approx\frac9{10}.$$
A: This is how the ancients might have attempted this, using the method of false position.
Solve $x^2=0.05$, starting with an initial guess of $x=0.2$.
$$0.05=(0.2+h)^2\approx0.04+0.4h\implies h=0.02$$
$$0.05=(0.22+h)^2\approx0.0484+0.44h\implies h=0.0036$$ So $x=0.2236$.
Now solve $y^3=0.2236$, starting with $0.6$.
$$0.2236=(0.6+h)^3\approx0.216+1.08h\implies h=0.007$$
$$0.2236=(0.607+h)^3\approx0.2236+1.1h\implies h=0$$
Finally solve $z^5=0.607$, starting with $0.9$.
$$0.607=(0.9+h)^5\approx0.5905+3.28h\implies h=0.005$$ so $z\approx0.905$.
Disclaimer: I used a calculator!
