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[EchoP]

This reminds of the fun I had messing with my new TI-89 Titanium today.

Did you know that calculators don't like it when you tell them to take quadnomials to powers of over a hundred. I like to play with equations like this. I just lack a calculator right now.

Press the on button if you want to stop it from working. It will break the operation.

[Virex]

2^x = 10^y

log_10(2^x) = log_10(10^y)

log_10(2^x) = y

Alternatively,

x = log_2(10^y)

Continuing from there you get X = Y*log_2(10) or Y=X*log_10(2) which you might recognise as a straight line. Now if you're looking for any point where X and Y are both integers, you'd need to find an X for which X * log_10(2) is an integer. That number would be
X = 1/(og_10(2) = log_10(10)/log_10(2) = log_2(10)
And since log_2(10) isn't an integer, X ins't an integer. So there's no pair of integers X and Y that fit this equation, and therefor
2^n = 10^m can only be true if M or N isn't an integer.

[Gantolandon]

2^n = 10^m can only be true if M or N isn't an integer.

2^0 = 1
10^0 = 1

Zero is definitely integer. Perhaps you meant "natural"?

[Virex]

2^n = 10^m can only be true if M or N isn't an integer.

2^0 = 1
10^0 = 1

Zero is definitely integer. Perhaps you meant "natural"?

Wups forgot about 0. Yes, 0 is the only integer that would fit the description. Any other integer (positive or negative) doesn't.

[Gantolandon]

After drawing graphs of both I think we could also safely rule out positive real numbers. I can't think a way these two could cross.

[Jim Groovester]

This isn't a difficult math problem. Virex already solved it a few posts ago, but it was a bit difficult to read. Hopefully, this helps.

2^x = 10^y
ln(2^x) = ln(10^y)
xln2 = yln10

y = x ln(2)/ln(10)

This is the solution curve to the original equation. This is a linear relation, and the Domain is (-∞,∞), and the Range is likewise (-∞,∞). There are no discontinuities. The trivial case of (0,0) is the only integer solution.

[Muz]

This isn't a difficult math problem. Virex already solved it a few posts ago, but it was a bit difficult to read. Hopefully, this helps.

2^x = 10^y
ln(2^x) = ln(10^y)
xln2 = yln10

y = x ln(2)/ln(10)

This is the solution curve to the original equation. This is a linear relation, and the Domain is (-∞,∞), and the Range is likewise (-∞,∞). There are no discontinuities. The trivial case of (0,0) is the only integer solution.

I give that answer a thumbs up. It contains little to no trolling elements, unlike some other answers

I'd go a bit further and say y = 0.3010x, because ln(2) and ln(10) are both constants.

[G-Flex]

If you wanted to keep it exact, you could simplify it to y = xlog₁₀2 as well

[Sowelu]

After drawing graphs of both I think we could also safely rule out positive real numbers. I can't think a way these two could cross.

You're doing the wrong thing there.

You shouldn't draw lines, you should just draw points at every integer along the horizontal axis.

Now, is there any pair of points on the horizontal line that give a matching pair of answers on the vertical axis?  They won't cross, but...

(I can't say x and y because they commandeered those...heh)