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: need to solve this question

  1. Dec 9, 2005 #1
  2. jcsd
  3. Dec 9, 2005 #2

    TD

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    Homework Helper

    With given y=ln(x) and y=ax it clear that for a certain x, ln(x) = ax. At least, we're looking for such an a.
    There is however a second condition in the point, (look at the hint) you can express that as: (ax)' = (ln(x))' <=> a = 1/x.
    Now substitute this a = 1/x in the first condition, this yields: ln(x) = 1. For what x does this hold? So what is a?
     
  4. Dec 9, 2005 #3

    Curious3141

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    Unfortunately, we cannot give you wholesale solutions, so I'll give you a hint.

    You need the line to touch the curve - that means that at that value of x, call it 'X' (big X), the y-coordinates are also equal. Come up with one equation to represent this.

    Next, you know that the gradient of the curve is the same as that of the line (which is a) at this point. Differentiate the curve to find the gradient at x = X. Now come up with another equation to represent this.

    Solve the two simultaneously for a, and you have your result.
     
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