# Fairly Simple Mathematics Question

**URL:** <https://boards.straightdope.com/t/fairly-simple-mathematics-question/161765>\
**Category:** Factual Questions\
**Created:** [March 16, 2003, 8:03am UTC](https://boards.straightdope.com/t/fairly-simple-mathematics-question/161765 "2003-03-16T08:03:53Z")\
**Posts on this page:** 3\
**Page:** 1

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**Author:** ![Netbrian](https://avatars.discourse-cdn.com/v4/letter/n/71e660/32.png) [@Netbrian](https://boards.straightdope.com/u/Netbrian)\
**Post date:** [March 16, 2003, 8:03am UTC](https://boards.straightdope.com/t/fairly-simple-mathematics-question/161765/1 "2003-03-16T08:03:53Z")

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Alright. On a whim, I started wondering whether there were any answers to the equation 2^x = 10^y for which x and y were both integers. I solved for y, and then wrote a QBASIC program to check every possible integer value of x and then tell me if it finds one that returns a y value.

So far, nothing.

So is it possible to tell whether or not this equation has a solution, and if so, can it tell what the solution is in a better way than guess/check?

Brian

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**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\
**Post date:** [March 16, 2003, 8:17am UTC](https://boards.straightdope.com/t/fairly-simple-mathematics-question/161765/2 "2003-03-16T08:17:05Z")

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The only solution in the integers is x = y = 0. The reason for this is that 2[sup]x[/sup] will never be a multiple of 5 for x \> 0, and 10[sup]y[/sup] will always be a multiple of 5 for y \> 0.

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**Author:** ![RM\_Mentock](https://avatars.discourse-cdn.com/v4/letter/r/e274bd/32.png) [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)\
**Post date:** [March 16, 2003, 4:06pm UTC](https://boards.straightdope.com/t/fairly-simple-mathematics-question/161765/3 "2003-03-16T16:06:28Z")

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Yep.

On the other hand, there are some that are close–which I assume prompted the question. x=10 and y=3 result in 1024=1000, which is close enough that the approximation is used all the time in the computer industry, a kilobyte being 1024 bytes for instance.

Taking the logarithms (base 10) of both sides, leaves  
log 2 = y/x  
This would be easy if the log of 2 were rational, but the log of 2 is not only irrational, it is transcendental, so there is no chance. However, log 2 = .3010299956, which explains why x=10 and y=3 work so well. Also, 2^100000 is pretty close to 10^30102, relatively. I have 'em as .99900209301438450794403276433e+30103 and 1e+30103.