# [Don't Merge][Prototype] Adding abs while converting equation to log form to get solved by _lambert #16960

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### jmig5776 commented Jun 4, 2019 • edited

#### Brief description of what is fixed or changed

This PR involves another method to fix the problem of bivariate.
This method involves:

• To take care of the missing logarithmic equations from the original equation.
• Then try to add proper solving method to handle the resulting abs included logarithmic equations after discussing with mentors.

#### Release Notes

NO ENTRY

``` Added abs while converting to logarithmic equations ```
``` 09536b4 ```

### sympy-bot commented Jun 4, 2019 • edited

 ✅ Hi, I am the SymPy bot (v147). I'm here to help you write a release notes entry. Please read the guide on how to write release notes. No release notes entry will be added for this pull request. Note: This comment will be updated with the latest check if you edit the pull request. You need to reload the page to see it. Click here to see the pull request description that was parsed. `````` #### References to other Issues or PRs #### Brief description of what is fixed or changed This PR involves another method to fix the problem of bivariate. This method involves: - To take care of the missing logarithmic equations from the original equation. - Then try to add proper solving method to handle the resulting abs included logarithmic equations after discussing with mentors. #### Other comments @Yathartha22 @aktech @Shekharrajak #### Release Notes NO ENTRY ``````
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### Motivation behind this method

Earlier some of the logarithmic equations were missed while converting the equations For eg :
While solving this equation `(a/x + exp(x/2)).diff(x) = 0`.

``````2*log(x) + x/2 - log(a) = 0 --> converted by sympy
2*log(-x) + x/2 - log(a) = 0 --> missed by sympy
``````

Now in this PR it considers both of the equation as

``````2*log(x) + x/2 - log(a) = 0 + 2*log(-x) + x/2 - log(a) = 0
\      /
\    /
\  /
2*log(abs(x)) + x/2 - log(a) = 0
``````

So by this all the logarithmic forms of original equations will be involved while solving.

### Problem with this method

Look at these following test cases failed and the equations involving in them.

```>>>solve(x*log(x) + 3*x + 1, x)
NotImplementedError: multiple generators [x, log(x)]
No algorithms are implemented to solve equation x*log(x) + 3*x + 1
# equation invoded in _lambert -> -log(_u) + log(_x*Abs(log(_x) + 3)) = 0
# clearly its very hard to solve this equation involving abs.

>>>solve(5*x - 1 + 3*exp(2 - 7*x), x)
NotImplementedError: multiple generators [x, exp(-x)]
No algorithms are implemented to solve equation 5*x + 3*exp(2 - 7*x) - 1
# equation involved in _lambert -> 7*_x + log(Abs(5*_x - 1)) - log(3*exp(2)) = 0

>>> solve((1/x + exp(x/2)).diff(x, 2), x)
[]
# equations involved log(exp(re(x)/2)*Abs(x**3)) - 3*log(2) = 0```

and many more examples with different errors.
Thats why I think #16890 is better beacuse at the end of both methods we had to consider every possible solution.

referenced this pull request Jun 4, 2019