MAKING SUBJECT OF FOMULAR

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lloyd mukunza
lloyd mukunza on 20 Sep 2014
Commented: Walter Roberson on 17 Dec 2018
HOW DO I MAKE X SUBJECT OF FOMULAR IN THIS EQUATION x = exp(x+y).
I have tried the following code
syms x y
solve('x = exp(x+y)',x)
and this give me *
[sym empty].*
How best can I do it.
  1 Comment
Chibuzo Chukwu
Chibuzo Chukwu on 13 Dec 2018
I am also stoked with a situation of the same form. All out gigs should please look into this kind of problem and suggest a way out.

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Answers (2)

Walter Roberson
Walter Roberson on 13 Dec 2018
In older versions of MATLAB,
solve('x = exp(x+y)', 'x')
In more modern versions
syms x y
solve(x == exp(x+y), x)
However, MATLAB is not able to find a solution. A solution exists, and is
-lambertw(-exp(y))
but MATLAB is not strong on Lambert W processing.
  1 Comment
Walter Roberson
Walter Roberson on 17 Dec 2018
Chibuzo Chukwu comments to me
Your contribution has been very useful

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Chibuzo Chukwu
Chibuzo Chukwu on 13 Dec 2018
Thanks for your response. I have tried to read about the Lambert W function and it has been helpful but can you look at this problem?
There are series of values for x and a and it's required to have a linear plot of a against x in the equation connecting them
ln(x^alpha * a) = c- 2*x*m
To determine values alpha and m from the graph
Is there anyway this can be rearrange to achieve this?
  15 Comments
Walter Roberson
Walter Roberson on 17 Dec 2018
No. If part of your data has a linear fit then it makes no sense to fit a power model to it.
If you want a linear fit then set alpha to 0. That would get ln a on the left side and linear terms on the right leading to aa simple polynomial fit.
Chibuzo Chukwu
Chibuzo Chukwu on 17 Dec 2018
Thanks a lot. Your insight has been very useful and infact you've answered my question very satisfactory. I will get back to you if I find any other obstacle. Thanks a million.

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