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Math Help - logs and e^x

  1. #1
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    logs and e^x

    Let M be a large real number. Explain briefly why there must be exactly one root A of the equation Mx = e^x with A > 1. Why is logM a reasonable approximation to A? Write
    A = logM + y. Can you give an approximation to y, and hence improve on logM as an approximation to A?

    Im not sure whether what I've done is correct:

    The graph of y=Mx and y=e^x only intersects once, as e^x is a much greater increasing function.

    M is a large number, so M=e^x approximately, so lnM = x approximately

    A = logM + y

    so y=logA

    Thanks
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  2. #2
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    Quote Originally Posted by Aquafina View Post
    Let M be a large real number. Explain briefly why there must be exactly one root A of the equation Mx = e^x with A > 1. Why is logM a reasonable approximation to A? Write
    A = logM + y. Can you give an approximation to y, and hence improve on logM as an approximation to A?

    Im not sure whether what I've done is correct:

    The graph of y=Mx and y=e^x only intersects once, as e^x is a much greater increasing function.
    You actually looked at the entire graph and not just a limited part? I don't believe that! What do you know about derivatives? Can you show that the derivative of e^x is larger than the derivative of Mx? You say "e^x is a much greater increasing function" but that is only true for very large x. You should be able to say exactly for what values of x the slope of Mx is larger than the slope of e^x and for what values of x it is less.

    M is a large number, so M=e^x approximately
    That doesn't even make sense. M is a constant and e^x depends on x. It certainly is NOT true if x is any small number!

    , so lnM = x approximately

    A = logM + y

    so y=logA
    NO! That would be saying "A= log M+ log A" and you have no reason to think that.

    Thanks
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    You actually looked at the entire graph and not just a limited part? I don't believe that! What do you know about derivatives? Can you show that the derivative of e^x is larger than the derivative of Mx? You say "e^x is a much greater increasing function" but that is only true for very large x. You should be able to say exactly for what values of x the slope of Mx is larger than the slope of e^x and for what values of x it is less.


    That doesn't even make sense. M is a constant and e^x depends on x. It certainly is NOT true if x is any small number!


    NO! That would be saying "A= log M+ log A" and you have no reason to think that.
    Ok so I have thought about this again, and now by thinking of the graph of y = Mx - e^x

    It has a maximum point, and 2 roots, 1 between 0 to 1 and the other which occurs later. So the root we are interested will be the later one.

    Am i right so far?

    Also, can you please give me a hint on the next bit, the logM approximation.
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