How close each other 2^x and 10^y can situate if x and y are natural numbers?

Jan 2013
20
0
Europe
If we have 2^3, it´s distance from 10^1 is only 2.
If we have 2^10, it´s distance from 10^3 is 24 (as 2^10=1024 and 10^3=1000).

Is there any other natural number pair x,y which results distance between 2^x and 10^y to be only 2?

Is there any formula, where to insert desired distance and find x and y as a result, or does it only requires raw computer power to find an answer?

(Wink)
 
Jan 2008
484
109
UK
I think it will never again be so close.

If on the other hand you consider 2^x / 10^y or vice versa you can get this close to 1.

For example \(\displaystyle 2^{196} \approx 1.004 \times 10^ {59}\) and this gives 59/196 as a good approximation of \(\displaystyle \log_{10}2\) for whatever that's worth. :)
 
  • Like
Reactions: 1 person

ILikeSerena

MHF Helper
Dec 2011
734
210
No, it won't be as close as 2.

For numbers \(\displaystyle x, y \ge 1\), we have:

\(\displaystyle 2^x - 10^y=\pm 2\)

\(\displaystyle 2^{x-1} - 2^{y-1}5^y=\pm 1\)

This can only be true if \(\displaystyle 2^{x-1}\) and \(\displaystyle 2^{y-1}5^y\) are relatively prime (Euclid).
That means that y-1=0.
So the only solution for a distance of 2 is: \(\displaystyle 2^3\) and \(\displaystyle 10^1\).
 
Last edited:
  • Like
Reactions: 2 people