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Tangent of logarithms

  1. Sep 15, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the equations of the tangents to the following graphs for the given values of x.

    (a) y = ln x, where x = 1/2

    2. Relevant equations



    3. The attempt at a solution

    I know ln x differentiated is 1/x but I cannot see when the rest fall into the place. The book I'm using doesn't explain it at all well.

    Answer: y = 2x - ln 2 - 1

    Thank you for looking :)
     
  2. jcsd
  3. Sep 15, 2010 #2

    danago

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    You have correctly identified the derivative of the function; you should then evaluate it at the point x=1/2 and remember that by the definition of the derivative, this will represent the SLOPE of the line tangent to the graph.

    Can you work from there?
     
  4. Sep 15, 2010 #3
    If you put 1/2 into 1/x you get 2.. so the tangent's gradient will be -2

    The logarithm's are really getting to me.. :/
     
  5. Sep 15, 2010 #4

    danago

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    Yes you get 2, but why would that make the gradient equal to -2?

    The equation for any general line is y=ax+b. You have just found the slope of this line, so it becomes:

    y = 2x + b

    How can you find b?
     
  6. Sep 15, 2010 #5
    Hmm.. I must be confusing myself with something else :\

    I'm following you so far (even though there's not much to follow lol).

    Okay, I'm unsure how to work b :S
     
  7. Sep 15, 2010 #6

    danago

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    Since we are finding the tangent to the curve at x=1/2, we know that the line must pass through the point with coordinates (1/2, ln(1/2)). If you substitute this into the linear equation that we have so far developed, it will define what b must be equal to:

    y = 2x + b
    ln (1/2) = 2(1/2) + b
    b = ln(1/2) - 1 = - ln(2) - 1

    Does that make sense?
     
  8. Sep 15, 2010 #7
    Ahh I understand it now - just these ln's cofusing me!

    Well thank you very much danago for your help :D
     
  9. Sep 15, 2010 #8

    danago

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    No worries :smile:
     
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