Simplifying the expression of exponential and logarithms I want to simplify the following expression.
$$Y=\text{Bottom} + \frac{\text{Top}-\text{Bottom}}{1+10^{((\log EC50-X))}}$$
$\log$ is base of $10$. Some may know that it's a dose response curve, and I want to solve for $EC50$.
I tried simplifying it but I forgot math long time ago and I couldn't find the answer online.
My problem was the exponential and logarithms part.
Can someone simplify the expression to solve for $EC50$?
Thank you very much.
Edit : And if someone could explain simplifying exponential and logarithm on this one, it would be great as well. I forgot all the math and I would appreciate it.
 A: Starting with $$Y = \text{Bottom} +\frac{\text{Top}-\text{Bottom}}{1+10^{\log_{10} EC50 -X}}$$
then using $10^{\log_{10} EC50 -X} = EC50 /10^X$ (if that is what you meant) you get $$Y -\text{Bottom} =\frac{\text{Top}-\text{Bottom}}{1+EC50 /10^X}$$ or $$1+EC50 /10^X =\frac{\text{Top}-\text{Bottom}}{Y -\text{Bottom}}$$ i.e $$EC50  =10^X\frac{\text{Top}-Y}{Y -\text{Bottom}}.$$
A: 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 $X$.
