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TomLikesPhysics

  • 2 years ago

If I have a function: y=K*e^x is there a way to rewrite it as y=C*A^x where A does not equal e?

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  1. experimentX
    • 2 years ago
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    let, A^i = e i = 1/lnA

  2. experimentX
    • 2 years ago
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    it will be the same nevertheless, however I am not sure about K and C

  3. TomLikesPhysics
    • 2 years ago
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    I´m just wondering if e is such a magical number to appear here and there or if its artificial and we could use a different set of numbers to express the same thing/physical law.

  4. experimentX
    • 2 years ago
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    i think e is quite an special number <-- especially when rate of change depends on initial value. also it has some special property (base of natural log), i think it is best to leave things with e's

  5. experimentX
    • 2 years ago
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    also stretching or compressing exponential function gives e at some point

  6. TomLikesPhysics
    • 2 years ago
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    Hmmm what is so special about e? I only know that the derivative of e^x is again e^x which is quite nice and that I can rewrite e^(ix) in terms of sine and cosine but is there even more to e?

  7. experimentX
    • 2 years ago
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    LOL ... not really sure!!

  8. TomLikesPhysics
    • 2 years ago
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    xD I thought there might be some additional stuff I might not know about.^^

  9. beginnersmind
    • 2 years ago
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    Medals 0 You can do it by writing e as a^(1/ln(a)) and apply exponential identities but why would you want to? e is such a nice number :)

  10. experimentX
    • 2 years ago
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    e's the best ... LOL

  11. beginnersmind
    • 2 years ago
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    From a practical point of view, if you need to differentiate or integrate your function down the road, would you rather deal with e^x or A^x?

  12. TomLikesPhysics
    • 2 years ago
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    @beginnersmind I was wondering if the appearance of e in some laws of physics was because somewhat had the hearts for e or because there is no other way to state that law. Might have been just some physicist who loved e and we could rewrite some laws. Ok than I guess e really is that great. :)

  13. experimentX
    • 2 years ago
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    comes naturally from \( \int 1/x dx \)

  14. TomLikesPhysics
    • 2 years ago
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    f I integrate or differentiate I am always happy to encounter e ;) But if I just add and multiply I could live without e ;) So it depends on the field.

  15. beginnersmind
    • 2 years ago
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    well, a lot of laws are actually of the form e^Cx, so I'm not sure e is especially preferred by nature in those cases. It seems to be notational .

  16. TomLikesPhysics
    • 2 years ago
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    So I could easily rewrite y=K*e^(Cx) with some other base?

  17. experimentX
    • 2 years ago
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    i've usually encountered in decay equation and distribution function. decay equation <--- dN/dt = N <-- depends on initial value distribution function --> didn't understand

  18. TomLikesPhysics
    • 2 years ago
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    I started wondering while looking at the decay equation like 10 Minutes ago.^^

  19. beginnersmind
    • 2 years ago
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    @experimentX in the decay function the choice of the base is somewhat arbitrary. Say you have f(t)=e^(-Ct). You could just as easily write f(t)-2^(-Kt), with a different constant.

  20. experimentX
    • 2 years ago
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    why ... we choose it as natural base for log while integrating <--- must be some reason.

  21. beginnersmind
    • 2 years ago
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    I'm saying e^(-Ct) and 2^(Ct/ln2) are the same function. They take the same values.

  22. beginnersmind
    • 2 years ago
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    2^(-Ct/ln2) that is

  23. TomLikesPhysics
    • 2 years ago
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    But If you rewrite it using the ln than e is still in there (in that function). So you can rewrite in a way that you can not see e but it is hidden in the ln.

  24. experimentX
    • 2 years ago
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    we could use the same logic everywhere, the point is why e so that there is no logs on the power?? 1/ ln 2 ??

  25. beginnersmind
    • 2 years ago
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    Well, there's a constant, which is an experimentally determined number. When you use e as the base the constant is C. When you use 2 it's K=C/ln2. If you actually measure half-life you're measuring K, so ln2 isn't really "hidden" there.

  26. TomLikesPhysics
    • 2 years ago
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    Ops right. ln2 is just some number - there is not an e hidden. So we could really rewrite equations that fit that pattern.

  27. experimentX
    • 2 years ago
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    not really sure if i am understanding ..:(

  28. TomLikesPhysics
    • 2 years ago
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    @experimentX Which part? You wrote the same thing as beginnersmind with the rewriting 1/lnA or now 1/ln2.

  29. experimentX
    • 2 years ago
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    e^x/nothing <--- why nothing in this case?? must be some special property of e A^x/lnA

  30. TomLikesPhysics
    • 2 years ago
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    Why e^x/nothing? Who wrote that where?

  31. experimentX
    • 2 years ago
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    usually equation comes that way when we integrate 1/x dx Oo, i think i need to review decay equation, since i ignored \( \lamda \) factor completely.

  32. experimentX
    • 2 years ago
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    LOL ... seems like only e is nice to deal with,

  33. TomLikesPhysics
    • 2 years ago
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    :) I guess than everything is alright. What a nice and interesting discussion. If you follow mathematics and physics in class it seems that e is mysteriously everywhere but apparently it is that way because some people are secretly working for e and we could write it in a different way. :) Nevertheless e is a great number for calculus.

  34. experimentX
    • 2 years ago
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    yeah ... that i agree!! makes nice, easy and clean.

  35. beginnersmind
    • 2 years ago
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    To be fair, there are situations where e appears in its own right. E.g. you can't rewrite e^i*pi=-1 with any number. (I think)

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