How many times do you need to double previous result to get at least $10^{82}$?

Question

This is pretty straightforward, but I'd like to study, how find out, how many times do you need to double previous result of calculation to get some sum, for example: $10^{82}$

1. $1\times 2 = 2$
2. $2\times 2 = 4$
3. $4\times 2 = 8$
4. $8\times 2 = 16$

n. $x\times 2 \geq 10^{82}$

Answer 1

The $n$th step is $2^n$. If you want $2^n\geq A$, then you want $$n = \log_2(2^n) \geq \log_2(A) = \frac{\ln(A)}{\ln(2)} = \frac{\log_{10}A}{\log_{10}(2)}.$$

So the first $n$ at which $2^n\geq A$ will be the least positive integer greater than or equal to $\log_2(A)$, which is denoted $$\left\lceil \log_2A \right\rceil.$$

Answer 2

For good ballpark estimates, use the fact that $2^{10}=1024\approx 1000$. So $10$ doublings is about the same as multiplying by $1000$. Since $10^{82}=10\times 1000^{27}$, we can see that $270$ doublings will get us past $10^{81}$. There is some slack, and another $3$ doublings get us over.