algorithms - Why does the method to find out log and cube roots work?

To find cube roots of any number with a simple calculator, the following method was given to us by our teacher, which is accurate to atleast one-tenths.

1)Take the number $X$, whose cube root needs to be found out, and take its square root 13 times (or 10 times) i.e. $\sqrt{\sqrt{\sqrt{\sqrt{....X}}}}$

2)next, subtract $1$, divide by $3$ (for cube root, and any number $n$ for $n$th root), add $1$.

3) Then square the resultant number (say $c$) 13times (or 10 times if you had taken out root 10 times) i.e. $c^{2^{2^{....2}}}=c^{2^{13}}$. This yields the answer.

I am not sure whether taking the square root and the squares is limited to 10/13 times, but what I know is this method does yield answers accurate to atleast one-tenths.

For finding the log, the method is similar:-

1)Take 13 times square root of the number, subtract 1, and multiply by $3558$. This yield s the answer.

Why do these methods work? What is the underlying principle behind
this?

Answer

Let's use these classical formulae :

$$e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$$
$$\ln\,x=\lim_{n\to\infty}n\left(x^{1/n}-1\right)$$

to get (replacing the limit by a large enough value of $n$ : $N=2^{13}$) :

$$\begin{align} \sqrt[3]{x}=e^{\left(\ln x/3\right)}&\approx \left(1+\frac {\ln x/3}N\right)^N\\ &\approx \left(1+\frac {N\left(x^{1/N}-1\right)}{3\,N}\right)^N\\ &\approx \left(1+\frac {\left(x^{1/N}-1\right)}3\right)^N\\ \end{align}$$

Concerning the decimal logarithm we have :
$$\log_{10}\,x=\frac{\ln\,x}{\ln\,10}\approx \frac N{\ln\,10}\left(x^{1/N}-1\right)$$

For $N=2^{13}$ we may (as indicated by peterwhy) approximate the fraction with $$\frac N{\ln\,10}=\frac {2^{13}}{\ln\,10}\approx 0.4343\times 8192\approx 3558$$

Hoping this clarified things,