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

• Jan 26th 2013, 02:18 AM
kpkkpk
How close each other 2^x and 10^y can situate if x and y are natural numbers?
If we have 2^3, itīs distance from 10^1 is only 2.
If we have 2^10, itīs 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 26th 2013, 07:35 AM
a tutor
Re: How close each other 2^x and 10^y can situate if x and y are natural numbers?
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 $2^{196} \approx 1.004 \times 10^ {59}$ and this gives 59/196 as a good approximation of $\log_{10}2$ for whatever that's worth. :)
• Jan 26th 2013, 07:57 AM
ILikeSerena
Re: How close each other 2^x and 10^y can situate if x and y are natural numbers?
No, it won't be as close as 2.

For numbers $x, y \ge 1$, we have:

$2^x - 10^y=\pm 2$

$2^{x-1} - 2^{y-1}5^y=\pm 1$

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