If you apply log transformations to all of your variables, the exponential function is used when you translate back the coefficients.

P.S.

The question is do you really need to translate back the coefs?

Logs are best for log-normal distributions and sqrt for Poisson distributions, see https://www.researchgate.net/post/Why_do_transformation_to_sqrt_is_better_than_log-linear_transformation

So if you obtained a good model (errors behave well), all you need to do is to transform back the predictions and you can draw graphs of how one variable depends on another variable.

conside a simple model like

y = b0 + b1*x1 + b2*x2 + error

(two predictors, x1 and x2, two "slope" coefficients b1 and b2, and the intercept b0).

Is you used log(y) instead of y, the model is

log(y) = b0 + b1*x1 + b2*x2 + error

what can be written as

y = exp(b0 + b1*x1 + b2*x2 + error)

   = exp(b0) * exp(b1*x1) * exp(b2*x2) * exp(error)

   = exp(b0) * exp(b1)^x1 * exp(b2)^x2 * exp(error)

so the exponentiated coefficients give you the base of the "fold-contribution" of the predictor. If the predictor is also used on the log-scale, the model -for instance with using log(x1) insted of x1- is

y = exp(b0 + b1*log(x1) + b2*x2 + error)

   = exp(b0) * exp(b1*log(x1)) * exp(b2*x2) * exp(error)

   = exp(b0) * x1^exp(b1) * exp(b2)^x2 * exp(error)

so the exponentiated coefficient is the power of the "fold-contribution".

For a square-root (sqr) transormation the interperation ca be difficult:

Give you fit

sqrt(y) = b0 + b1*x1 + b2*x2 + error

the original response is

y = (b0 + b1*x1 + b2*x2 + error)^2

=  b0^2 + 2*b0*b1*x1 + 2*b0*b2*x2 +2*b0*error + b1^2*x1^2 + 2*b1*b2*x1*x2 + 2*b1*x1*error + b2^2*x2^2 + 2*b2*x2*error + error^2

So here the coefficients are not easy to interpret because ther contributions can not be isolated.

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I would not recommend sqrt-transformation for Poisson-distributed data. Instead you should better use a Poisson regression model (usually employing a log-link function).

About the prev answer: sqrt was used for independent vars and logs for the outcome. So the Poisson regression seems to be not the best solution.

Negative binomial regression is more flexible, so could be probably a solution here.

If errors behave well after transformations (are normally distributed), I would not recommend anything else - just provide a graph to show how the dependency looks like.

Ah, yes, sqrt was used only for predictos. That's no problem.The interpretation of the coefficient, call it "b", remains straight-forward: per one unit increase of sqrt(x) does the expected response change by b units.

I suggested the Poisson regression as a substitute for a sqrt-transformation of the response when this is count data. You are right that the negative binomial regression is more flexible (it can handle over-disperision), but this also refers to count data as the response. Since Mehdi did not and sqrt-transformation of the response, both recommendations (Poisson and neg.binom.) to not fit his problem and can be ignored.

I agree to your last statement, when we consider specifically the log-transformation of the response. Other transformations can also give nice-behaving residuals but they often render the interpretation of the model (and of the coefficients) impossible*, what I think is not a good way to go. The log-transformation (of the response) often makes a lot of sense, and it changes an additive model into a multiplicative model (so not changes are not "valued" by their absolute size but by ther relative size!), so that effects are proportional to the expected value of the response.

* a counter-example is the Lineweaver-Burk transformation (double reciprocal) where intercept and slope do have in fact well-interpretable meanings (see https://en.wikipedia.org/wiki/Lineweaver%E2%80%93Burk_plot).

It would be useful here to know what kind of data is modelled.

Is it count data?

Thank you very much Andrey and Jochen for this valuable discussion. I am learning a lot already reading your comments line by line. The response variable is EQ5D which is a score (0-100) given by participants. Its distribution is very strange as said by one of my expert colleagues. Its range fits between 20 to 100 however there are very few observations below 50. The independent variables are age at enrollment, age at treatment and BMI for patients older than 15 years. Again the distribution of these variables were also not so fit for a linear regression. I had problem with linearity and the hetroscedasticity. Therefore I used these transformations.  

Generally, "scores" are actually ordinal scaled variables, and, generally, linear models (regression, anova, ancova, t-test...) require interval scaled variables. However, a score variable with 101 levels might possibly be interpreted as "quasi-interval" scales variable.

However, what is more of a concern for me is that the score sclae is bounded: negative values and values above 100 are not possible. This explains much of your findings that the relation is not linear and the variances are not homogeneous.

One way to solve this is to transform the scale values (y) into the range [0;1] by y'=(y/101)+1/(2*101),* and use the resulting values y' to fit a beta-binomial regression model. A more crude way (a workaround) is to logit-transform the y' values: y'' = log( y' / (1-y') ) and go for a linear regression of y''.

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* The simple way to get y' is y'=y/100, but this would give you y' values of 0.0 and 1.0, which can not be used subsequently. The formula presented above sqeezed the range of possible values for y' from 0.005 to 0.995.

Thank you very much Dear Jochen. I really appreciate your time and valuable guidance. Best regards, Mehdi

Btw, for those who are still interested:

if y is in percentages and is between 0 and 100, you can apply logit or arcsin+sqrt (google it) for y/100.