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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.

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  • $\begingroup$ To start you off: $10^{\log_{10} x} = x = \log_{10} 10^x$. Also $\log a + \log b = \log(ab)$ and $\log c - \log d = \log(c/d)$ in the same way as $10^{g+h} = 10^g \times 10^h$ and $10^{j-k} = 10^j / 10^k$. $\endgroup$
    – Henry
    Jun 21 '12 at 21:58
  • $\begingroup$ Does LogEC50-X mean $\log(EC50)-X$ or $\log(EC50-X)$? $\endgroup$
    – Henry
    Jun 21 '12 at 21:59
  • $\begingroup$ Since its 10^$\log EC50-X$, it wouldn't really make sense if it wasn't $\log(EC50-X)$; otherwise it would simplify immediately. $\endgroup$ Jun 21 '12 at 22:15
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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 $$EC50 =10^X\frac{\text{Top}-Y}{Y -\text{Bottom}}.$$

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  • $\begingroup$ Thank you so much. I just didn't remember how minus worked when it is on the exponentials. It was very helpful. Thank you! $\endgroup$
    – js0823
    Jun 22 '12 at 1:25
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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.

  1. $10^{\log a}=a$

  2. $\log(10^b)=b$.

  3. $\log(ab)=\log a+\log b$

  4. $\log 1=0$; $\log(1/a)=-\log a$

  5. $\log(a^c)=c\log a$

  6. $10^{a+b}=(10^a)(10^b)$

  7. $10^{ab}=(10^a)^b$.

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 $X$.

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  • $\begingroup$ Thank you so much for your answers. It seems I made everyone confused about the exponentials and logarithms. Your remark is correct. Your general tips were very helpful as well. I wish I could choose more than one correct answers. Thank you! $\endgroup$
    – js0823
    Jun 22 '12 at 1:24

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