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    Why is the inverse of  e^x = log_eX and not  log_ee^x.
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    From a general theory point of view, an inverse function is such that f^{-1}(f(x))=x

    so if f(x)=e^x then the inverse function is log_{e}(x) since then f^{-1}(f(x))=log_{e}(e^x)=x
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    log(e)e^x = xlog(e)e = x.1 = x
    This is clearly not the inverse of e^x!

    Sketch e^x and lnx - you'll see that they are a reflection in y = x, which means that they are the inverse of each other.
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    (Original post by AnonyMatt)
    log(e)e^x = xlog(e)e = x.1 = x
    This is clearly not the inverse of e^x!

    Sketch e^x and lnx - you'll see that they are a reflection in y = x, which means that they are the inverse of each other.
    thats true, im still reading the second post, although after i went away to think i realised that  log_ee^x can't be the inverse.
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    (Original post by ForGreatJustice)
    From a general theory point of view, an inverse function is such that f^{-1}(f(x))=x

    so if f(x)=e^x then the inverse function is log_{e}(x) since then f^{-1}(f(x))=log_{e}(e^x)=x
     f^-^1(f(x))=x? How is this the same as f^-^1(x)=x and is it meant to be?
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    (Original post by Core)
    How is this the same as f^-^1(x)=x and is it meant to be?
    It's not the same, and it's not meant to be.
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    (Original post by ForGreatJustice)
    It's not the same, and it's not meant to be.
    thankyou i was trying to make it the same in my mind and it was confusing me.  f^-^1(f(x))=x = f^0(x)=1(x)=x?
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    So this is from the functions chapter isn't it, So if I try to do the opposite to e^x I will end up at  log_ex ok then
    how about this  f(x)=e^x, f^-^1(x)=log_ex, ff^-^1(x)=x , opps i dont think ive dealt with functions that turn x into a power of something, but il think it over in my head and try to make sense of it, il post my thoughts, please say if they are correct.
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    Apply  f(x)= e^x, ff^-^1(x)=x so the  f^-^1(x) is a function that maps  e^x to  x alright i get it now.  ff^-^1(x)=x because  f(x)=e^x making  f^-^1(x) = f^-^1(e^x) in  ff^-^1(x) and  log_ee^x = x as log  log_ex is the function of  f^-^1 as to why this function qualifies as the inverse, is it a case of as long the I use function when acting on another function gets me back to x then this function (the one being applied) is the inverse of the other function (the one being acted upon)?
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    (Original post by Core)
     f^-^1(x) = f^-^1(e^x)

    This isn't what you mean, surely?

    But the rest of the post seems fine, I think.
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    (Original post by ForGreatJustice)
    This isn't what you mean, surely?

    But the rest of the post seems fine, I think.
    what i meant by that was in the composite function ff^-^1(x)=x
    after applying  f(x)=e^x you then apply  f^-^1(x)=log_ex only in the case of the composite function  x becomex  e^x
    so you have instead of  f^-^1(x)=log_ex we have  f^-^1(e^x)=log_ee^x but this is only in the case of f(x)=e^x.
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    I read your post, but I don't actually know what you're trying to say, or if you even saying anything. The point is the composition of a fucntion and it's inverse gives you the input argument, in this case it is x.
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    (Original post by ForGreatJustice)
    I read your post, but I don't actually know what you're trying to say, or if you even saying anything. The point is the composition of a fucntion and it's inverse gives you the input argument, in this case it is x.
    The point of the post was to explain to you what i meant by  f^-^1(e^x) i was saying that this was only in the second stage of the composite function  ff^-^1(x)=x, i wanted to clear up any confusion you had about my post.
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    Ok, well f^{-1}(e^x)=log_e (e^x)=x

    Are you fine with it now, or is anything still a problem?
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    (Original post by ForGreatJustice)
    Ok, well f^{-1}(e^x)=log_e (e^x)=x

    Are you fine with it now, or is anything still a problem?
    No im fine with it now thank you.
 
 
 
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