Just a general question....

[rsbriggs]

About sizes of numbers in general.

Is there a formula to calculate, say, the maximum x and minimum y such that 2^x has less than some arbitrary number of digits, say ten-million, and 2^y has more than ten-million digits?

Curious people want to know.....

[nuutti]

lenght of the number x is log10(x)

when we have 2^x -> log10(2^x) = x*log10(2) = x*0.301029996


from high school math.

Yours,

Nuutti

[rsbriggs]

Unfortunately, my last high school math class was in the general vicinity of 45 years ago...

So do I understand this to say that X would need to be in the general vicinity of 3 million to generate a million digit number for 2^X ? Which would mean that the exponents we are testing are starting to push into the 450,000 digit range?

[Hawk]

Roughly, multiply by 0.3 to go from X to number of digits, and divide by 0.3 to go from digits to X.
So we are currently at X = 1 400 000, which has 420 000 digits and for 1 million digits you need X around 3 300 000.