Solving an equation

Question

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.

Answer 1

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

Answer 2

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

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

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

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.