You can write an equation for a dose response curve either in terms of EC50 or log(EC50). Curve fitting finds the curve that minimizes the sum-of-squares of the vertical distance from the points. Rewriting the equation to change between EC50 and log(EC50) isn't going to make a different curve fit better. All it does is change the way that the best-fit EC50 is reported. However, rewriting the equation to change between EC50 and log(EC50) has a major effect on standard error and confidence interval of the best-fit values. Consider these sample results:

These data were fit to a dose-response curve with a standard slope. The best-fit value for logEC50 is -6.059.  Converting to the EC50 is no problem - simply take the antilog. The EC50 is 10^-6.059 M, about 0.87 mM.

The standard error of the logEC50 is 0.2717. It is used as an intermediate result to calculate a confidence interval, which ranges from -6.657 to -5.461.  This means that the 95%CI of the EC50 extends from 10^-6.657 to 10^5.461 -- from 0.2207 to 3.458 mM. Expressed as concentrations (rather than log of concentration) the interval is not centered on the best-fit value (0.87 mM). Switching from linear to log scale turned a symmetrical confidence interval into a very asymmetrical interval, which you can report.

If you fit the same data to an equation describing a dose-response curve in terms of the EC50 rather than the logEC50, the EC50, remains 0.87 mM.  But now the program computes the SE of the EC50 (0.5459 mM), and uses this to compute the 95% confidence interval of the EC50, which ranges from 0.3290 to +2.074 mM. Note that the lower limit of the confidence interval is negative! Even setting aside the negative portion of the confidence interval, it includes all values from zero on up, which isn't terribly useful. The uncertainty of the EC50 really isn't symmetrical, so the confidence intervals are not useful if you fit to a model written in terms of the EC50.

When some people see the SE of the logEC50, they are tempted to convert this to the standard error of the EC50 by taking the antilog. In the example, the SE of the logEC50 is 0.2717. The antilog of 0.2717 equals 10^0.2717 or 1.896. What does this mean? It certainly is NOT the SE of the EC50. The SE does not represent a point on the axis; rather it represents a distance along the axis. A distance along a log axis does not represent a consistent distance along a linear (standard) axis. For example, increasing the logEC50 1 unit from -9 to -8 increases the EC50 9nM; increasing the logEC50 1 unit from -3 to -2 increases the EC50 by 9 mM (which equals 9,000,000 nM). So you cannot interpret the number 1.896 as a concentration. You can interpret it as a multiplier - a factor you multiply by or divide into the EC50.  To calculate the 95% CI, first multiply 1.896 by a constant from the t distribution for 95% confidence and the appropriate number of degrees of freedom (2.2201 for this example). The result is 3.963. Then compute the 95% CI of the EC50. It extends from the best-fit EC50 divided by 3.963 to the best-fit EC50 times 3.963, from 0.22 mM to 3.45 mM.