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I have a regression as follows:

ln(y + 5) = A + B*ln(x + 5)

The addition of the number 5 in both logarithms is done simply because there are negative numbers in both variables, but I would like to know what consequences this addition has on the slope B and if I could calculate from this slope B the slope it would have had if the regression had been of the form:

ln(y) = A + B*ln(x)

How could the addition of 5 be undone? How does this addition affect the slope B?

EDITED:

My problem is the following:

What I want to do is a sensitivity exercise for variable Y based on variables A and B. The model we have developed only relates Y to A so for different percentages (variations of A) we can take out the results of Y. Now we want to do the same but varying the variable B, i.e. vary the variable B by % and see how Y varies (the problem is that we have no model that relates them and therefore the idea is to create a relationship between A and B that relates the increments so that if my variable B increases by 10% I know that corresponds to a -5% of A and then I go to my results table and get those values of Y).

For simplicity and to understand each other we can assume that A is the Annual Variation of the Unemployment Rate and B the annual variation of the GDP. My understanding of the problem tells me that in order to translate from % to % I have to do the regression between these two variables using the logarithms (so that the meaning of the slope is: if I increase the GDP 1% the Unemployment increases B%), and the problem is that this is where I introduce the +5 because these variables have negative numbers...But maybe I am interpreting something wrongly.

Alberto Gonzalez Almuiña
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    This does not totally make sense to me. As you (correctly) point out, the logarithm does not make sense for values of zero or lower, so what do you mean “if the regression had be of the form” that it cannot be? – Dave Sep 24 '21 at 11:40
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    Because the relationship is not linear this, what you are suggesting, cannot be done. The results will show a different picture, which won't answer your question. It would be better if you explain to us what is your actual problem, see https://meta.stackexchange.com/questions/66377/what-is-the-xy-problem. – user2974951 Sep 24 '21 at 12:38
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    The model with the "start value" of $5$ and the model without it are not linearly equivalent. The slope of one translates (if it's at all possible) into a *curvilinear* relationship in the other. For more about adding start values for logarithms, see https://stats.stackexchange.com/questions/30728. – whuber Sep 24 '21 at 14:18
  • I have edited the question with a much broader description of the problem because I believe that this way my problem will be much better understood and not so much the search for a concrete solution, so as not to fall into the XY problem. – Alberto Gonzalez Almuiña Sep 27 '21 at 08:06

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