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Integrating gives $$\ln\frac{250-X}{40-X} = 210kt+c_1\qquad\text{or}\qquad \frac{250-X}{40-X}=c_2e^{210kt}.\tag{10}$$ When $t=0, X=0,$ so it follows at this point that $c_2 =\frac{25}{4}$. Using $X=30g$ at $t=10$, we find $210k=\frac{1}{10}\ln\frac{88}{25} = 0.1258.$ With this information we solve the las equation in $(10)$ for $X$: $$X(t) = 1000\frac{1-e^{-0.1258t}}{25-4e^{-0.1258t}}.\tag{11}$$

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Believe me, I've tried so hard and I cannot get equation #11.

MJD
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Javittoxs
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  • It would be more helpful if you provide your work to me instead of showing that you "tried so hard". It is not enough to determine which part you are stuck on (except that you can't get to equation 11). – NasuSama Mar 10 '14 at 00:49

2 Answers2

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Try calling $c_2 e^{210kt}$ something like $K$, and the equation will look less scary and try solving that. Then plug back in. Hope this helps, Trurl

Trurl
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$(a-x)/(b-x)=ke^{ct},$

$a-x= bke^{ct}-xke^{ct},$

$x(1-ke^{ct})=a-bke^{ct}.$

Now divide through by $1-ke^{ct}$ and you have solved for $x$.

Of course you have to plug in what $a,b,k,c$ are from your specific equation.

Claude Leibovici
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coffeemath
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