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\noindent (``Simple chemical conversion") From chemical experiments, it is known that for certain reactions where a substance converts into another substance, the rate of change (with respect to time) of the amount $x$ at any time $t>0$ is determined by the differential equation
$$\dfrac{\mathrm{d}x}{\mathrm{d}t}=-kx(t),$$
for some constant $k$ (we use ``$-k$" in the equation because the amount of the substance is decreasing as it converts). The initial amount of the substance at time $t=0$ is $x(0)=x_0$.
\noindent Suppose also that it is known (by measurement) that half of the substance converts by the end of $10$ seconds.
\noindent At what time does $\dfrac{9}{10}$ of the substance become converted? \\
\textit{Solution}: Solve the differential equation by ``separation of variables", i.e. ``illegally" splitting up the $\dfrac{\mathrm{d}x}{\mathrm{d}t}$ by ``multiplying" by $\mathrm{d}x$ to get
$$\mathrm{d}x = -k x \mathrm{d}t.$$
Divide both sides of this equation by $x$ (so that the $x$ is on the same side as $\mathrm{d}x$) to get
$$\dfrac{1}{x} \mathrm{d}x = -k \mathrm{d}t.$$
(Illegally) integrate both sides by placing a $\displaystyle\int$ symbol in the front of each side:
$$\displaystyle\int \dfrac{1}{x} \mathrm{d}x = \displaystyle\int -k \mathrm{d}t.$$
Integrate on both sides (recall that the constant $-k$ can pull outside of the integral) to get
$$\ln(x) = -kt+C.$$
To solve for the function $x$, plug both sides into the exponential function to get the function
$$x(t) = e^{-kt+C} = e^C e^{-kt} = C_1 e^{-kt}, \quad C_1=e^C.$$
\underline{Find $C_1$} \\
Using the initial condition $x(0)=x_0$, we observe that
$$\underbrace{x_0}_{\mathrm{given}}=x(0)=\underbrace{C_1 e^{-k0}=C_1e^0 = C_1}_{\mathrm{calculated}}$$
Therefore $C_1=x_0$. This means our model is
$$x(t) = x_0 e^{-kt}.$$
\underline{Find $k$} \\
To find $k$ we use the statement that half of the substance is gone at time $t=10$ to write
$$\underbrace{\dfrac{1}{2}x_0}_{\mathrm{given}}=x(10)=\underbrace{x_0 e^{-10k}}_{\mathrm{calculated}}.$$
Now solve this for $k$: divide both sides by $x_0$ to get
$$\dfrac{1}{2} = e^{-10k}.$$
Take $\ln$ of both sides to get
$$\ln \left( \dfrac{1}{2} \right) = -10k,$$
and finally divide by $-10$ to get
$$k=-\dfrac{1}{10}\ln \left( \dfrac{1}{2} \right) = \dfrac{\ln(2)}{10}=0.0693147\ldots .$$
Therefore our model is
$$x(t) = x_0 e^{-\frac{\ln(2)}{10}t}.$$
To answer the question ``At what time does $\dfrac{9}{10}$ of the substance become converted?" we must be careful in setting up the equation. If $\dfrac{9}{10}$ of the subtsance has converted, then how much is left? The answer is $\dfrac{1}{10}$! Therefore we want to solve for $t$ in the following equation:
$$\dfrac{1}{10}x_0 = x(t) = x_0 e^{-\frac{\ln(2)}{10}t}.$$
Dividing by $x_0$ yields
$$\dfrac{1}{10} = x(t) = e^{-\frac{\ln(2)}{10}t}.$$
Taking $\ln$ of both sides yields
$$\ln \left( \dfrac{1}{10} \right) = -\dfrac{\ln(2)}{10}t.$$
Hence
$$t=-\dfrac{10}{\ln(2)} \ln \left( \dfrac{1}{10} \right) = \dfrac{10\ln(10)}{\ln(2)}=33.219280\ldots .$$
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