How many times must you square a number to get $<1/2$

Question

Let $0\leq x<1$. Be given. How many times must you square $x$ to get less than $1/2$?

Clearly this depends on $x$. But is there a nice formula to determine this? Such as: To make $1-1/n<1/2$, square it $n!$ times.

Answer 1

Suppose $m$ satisfies $x^m < \frac{1}{2}$. Since $\log_2$ is an increasing function, we have $$m \log_2 x = \log_2(x^m) < \log_2 \left(\frac{1}{2}\right) = - 1.$$ Since $x < 1$ we have $\log_2 x < 0$ and so dividing through by $\log_2 x$ gives $$m > -\frac{1}{\log_2 x}.$$

On the other time, squaring a number $x$ n times gives $(\cdots (x^2)^2 \cdots)^2 = x^{2^n}$, so if squaring $x$ $n$ times gives a number smaller than $\frac{1}{2}$ we must have $$2^n > - \frac{1}{\log_2 x},$$ that is, $$n > \log_2 \left( - \frac{1}{\log_2 x} \right).$$ Since we can only square $x$ an integer number of times, the minimum number of squarings is the smallest integer larger than the right-hand side, namely $$\left\lfloor \log_2 \left( - \frac{1}{\log_2 x} \right)\right\rfloor + 1,$$ which is also the number of digits before the decimal point of $- \frac{1}{\log_2 x}$.