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Plot inverse function Mathematica

  1. Mar 5, 2010 #1
    I want to plot inverse function using Mathematica
    Code (Text):

    In[1]:=f = Solve[x == a * Log[y/100], y]

    Out[2] = {{y -> 100 E^(x/a)}}

    and then? how to use this reult to plot?


    Another question is about the axeslabel, when I set the Frame->True, the AxesLabel can not be displayed correctly, where the label of x-coordinate is correct, but that of y-coordinate is missing, How to solve this problem?

    thanks again
    Last edited: Mar 5, 2010
  2. jcsd
  3. Mar 5, 2010 #2


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    Gold Member

    You can strip out the result from the brackets:

    Code (Text):

    In[209]:= f = Solve[x == a*Log[y/100], y]
    Out[209]= {{y->100 E^(x/a)}}
    In[212]:= F = f[[1]][[1]][[2]]
    Out[212]= 100 E^(x/a)
    Then Plot[F,{x,-3,3}] to plot. (assuming you have a value for "a")
  4. Mar 5, 2010 #3


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    Gold Member

    Also, once you use "Frame-> True" you need to change your "AxisLabel" to a "FrameLabel". That should solve it.
  5. Apr 5, 2010 #4

    I am completely a beginner therefore I would like to ask what you did in the In[212].. i mean what is the philosphy behind it..

    Thank you in advance

  6. Apr 6, 2010 #5


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    Gold Member

    209: I set f is the solution set {{y->100 ...}} etc
    212: I basically say, take the 2nd component of the 1st component of the 1st component of "f"

    So you have :
    {{y->100 E^(x/a)}}[[1]] = {y->100 E^(x/a)}
    {{y->100 E^(x/a)}}[[1]][[1]] = y->100 E^(x/a)
    {{y->100 E^(x/a)}}[[1]][[1]][[2]] = 100 E^(x/a)

    Basically the {}{} brackets are like a set, or table. And since its doubly thick {{X-> Y}} You need to pull out "Y".
    the [[1]] and [[2]] takes the first or second or whatever value of the set preceeding it.
    {1,2,3,4,3,2,1}[[5]] = 3 (indexed the 5th component)

    If there were two solutions you would get to choose. Notice:

    In[11]:= f = Solve[x^2 == 4, x]
    Out[11]= {{x->-2},{x->2}}
    In[12]:= f[[1]]
    Out[12]= {x->-2}
    In[13]:= f[[2]]
    Out[13]= {x->2}
  7. Apr 6, 2010 #6

    Great! Just on time since I was trying to do exaclty what u illustrate in your last example..thank u very much!!!!
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