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Y=ln(3-x) How to transform it? What is wrong with my logic?

  1. Sep 25, 2012 #1
    Hi, my name is Tim, I am new to this forum. Here's the description:

    I don't understand how they've done it. I realise I can plot this, but what is the logic.

    As a graph, I drew ln(-x) first, then added 3, so move left by 3 but that was not correct
    I can see they've added 3 first, move left 3. Then they minus 1 times x. Or have they done it another way?
    Why wasn't the first one correct, it makes sense to me

    I also tried another approach which was ln(-(x-3)) so I first drew ln(x-3), and then I minus 1 times x, so I flipped it against the y axis. That also did not work, what is wrong with the logic on this one as well?
  2. jcsd
  3. Sep 25, 2012 #2


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    Science Advisor
    Homework Helper

    I really don't understand. The graph of [itex] \ln(3-x) [/itex] is the curve which has exactly the same shape as [itex] \ln x [/itex], only that it is "crosses" the Ox axis in x=2 and diverges rapidly to [itex] -\infty [/itex] in x=3.

    In other terms, it the graph of [itex] \ln -x [/itex] (which you get from the graph of [itex] \ln x [/itex] by rotating it around the Oy axis) shifted rightwards on the Ox axis by 3 units.
  4. Sep 26, 2012 #3
    Thanks for the reply. Can you please explain/reword

    I'm looking at it as transformation, so just like y=(x^2 + 3) moves x^2 up 3 units. Why isn't y=ln(3-x) moving ln(-x) 3 times to the left? and also to the last part of my question of my first post, if you write it as ln(-(x-3)) why wouldn't that work?

    much appreciate for the help
  5. Sep 26, 2012 #4


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    Staff: Mentor

    Why not? Because ln(3-x) is not the same as (ln(-x)) - 3

    You should be able to show, viewing it as the sum of two logarithms,
    ln(-x) - 3 = ln( (-x) / (e³) )
  6. Sep 26, 2012 #5


    Staff: Mentor

    Here's why.

    Basic function: y = ln(x)
    Reflection across y-axis: y = ln(-x)
    Translation: y = ln(-x + 3) = ln(-(x - 3))
    Notice that this is a translation to the right by 3 units of the graph of y = ln(-x).
    Draw the basic untransformed graph, y = ln(x)
    Reflect across the y-axis to get y = ln(-x)
    Translate right by 3 units to get y = ln(-(x - 3)

    On the first graph, one point is (1, 0).
    On the reflected graph, this point moves to (-1, 0) [Check: ln(-(-1)) = 0]
    On the translated graph, this point (in previous step) moves to (2, 0) [Check: ln(-(2 - 3)) = ln(-(-1)) = ln(1) = 0.
  7. Sep 28, 2012 #6
    Thanks for the responses.

    After reading mark's reply its starting to clear up more. Much appreciate the help :)
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