Mathematica Plot inverse function Mathematica

[meridian]

I want to plot inverse function using Mathematica

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?

ThanksAnother 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

 

[Hepth]

You can strip out the result from the brackets:

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")

 

[Hepth]

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

 

[Yannis]

You can strip out the result from the brackets:

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")

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

Yannis

 

[Hepth]

ok:
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}

 

[Yannis]

ok:
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}

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