Please see the remark at the end, since I don't know for sure what you intend by your formula.
I will use somewhat different notation for ease of typing. Write $w$ for EC$50$, $t$ for top, $b$ for bottom. If I interpret things right, you have $Y=b+\frac{t-b}{1+10^{\log(w-X)}}.$
The part you are worried about is the simplest, for in general $10^{\log a}=a$. So your equation can be rewritten as $Y=b+\frac{t-b}{1+w-X}.$ the rest is just algebra. First we have $Y-b=\frac{t-b}{1+w-X}.$ Flip both parts over. We get $\frac{1}{Y-b}=\frac{1+w-X}{t-b},$ and then, multiplying both sides by $t-b$, we get $\frac{t-b}{Y-b}=1+w-X.$ Extracting $w$ is now easy. We get $w=\frac{t-b}{Y-b}-1+X.$ You may wish to use the simplification $\frac{t-b}{Y-b}-1=\frac{t-Y}{Y-b}$.
About general tips, here is a short list of what you need to know.
$10^{\log a}=a$
$\log(10^b)=b$.
$\log(ab)=\log a+\log b$
$\log 1=0$; $\log(1/a)=-\log a$
$\log(a^c)=c\log a$
$10^{a+b}=(10^a)(10^b)$
$10^{ab}=(10^a)^b$.
If later you ask a specific question or questions you have thought about, someone on this site can show you how to handle the computations.
Remark: If what is intended in the formula is $10^{\log(w)-X}$, this simplifies to $(10^{\log w})(10^{-X})$, and then to $\frac{w}{10^X}$. The rest of the manipulations until close to the end are the same. But ultimately to get $w$ we multiply by $10^X$ instead of adding $X$.