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

#### kpkkpk

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)

#### a tutor

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.

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#### ILikeSerena

MHF Helper
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 $$\displaystyle 10^1$$.

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