44=2log(E/2)+logE

Oops, I didn't notice you'd changed to log base 10 in the question, making it different to what was in the subject.

Are these supposed to both be logs to base 10, or only one? It doesn't change the method, anyway.

I think you'll be able to work it out for yourself once I show you that the equation that gave rise to what you wrote is this:

$${{\mathtt{e}}}^{{\mathtt{44}}} = {\left({\frac{{\mathtt{E}}}{{\mathtt{2}}}}\right)}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{E}}$$

Take logs of both sides, and you arrive at what you began the question with. So all you have to do is solve for E in the equation I wrote.

Easy enough?