How do I solve this simultaneous equation that has the constant $e$ inside?

Question

$28.8=24.5+Ce^{(-kt)}$ -(1)

$28.0=24.5+Ce^{-k(t+ \frac{29}{60})}$ -(2)

What I did so far:

$24.5=28.8-Ce^{(-kt)}$

$24.5=28.0-Ce^{-k(t+ \frac{29}{60})}$

$28.8-Ce^{(-kt)}=28.0-Ce^{-k(t+ \frac{29}{60})}$

$0.8-Ce^{(-kt)}=-Ce^{-k(t+ \frac{29}{60})}$

$Ce^{(-kt)}-0.8=Ce^{-k(t+ \frac{29}{60})}$

And I'm stuck at this part cause I can't use $log_e$ as the other side will be affected as well...

Answer 1

We want to solve $\text{k}$ for a real value, for a given real value of $\text{a},\text{b},\text{x},\text{y}$ and $\text{z}$:

$$ \begin{cases} \text{a}=\text{b}+\text{C}e^{-\text{k}t}\\ \text{x}=\text{y}+\text{C}e^{-\text{k}\left(t+\text{z}\right)} \end{cases} $$

Using:

• $$\exp[\ln(q)]=e^{\ln(q)}=q$$
• $$\exp[n+m]=e^{n+m}=e^n\cdot e^m$$

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So, we get:

$$ \begin{cases} \text{a}=\text{b}+\text{C}e^{-\text{k}t}\\ \text{x}=\text{y}+\text{C}e^{-\text{k}\left(t+\text{z}\right)} \end{cases}\Longleftrightarrow \begin{cases} \text{a}-\text{b}=\text{C}e^{-\text{k}t}\\ \text{x}-\text{y}=\text{C}e^{-\text{k}\left(t+\text{z}\right)} \end{cases}\Longleftrightarrow $$ $$ \begin{cases} \frac{\text{a}-\text{b}}{\text{C}}=e^{-\text{k}t}\\ \frac{\text{x}-\text{y}}{\text{C}}=e^{-\text{k}\left(t+\text{z}\right)} \end{cases}\Longleftrightarrow \begin{cases} \ln\left[\frac{\text{a}-\text{b}}{\text{C}}\right]=-\text{k}t\\ \ln\left[\frac{\text{x}-\text{y}}{\text{C}}\right]+\text{k}\text{z}=-\text{k}t \end{cases} $$

So, we get:

$$\ln\left[\frac{\text{a}-\text{b}}{\text{C}}\right]=\ln\left[\frac{\text{x}-\text{y}}{\text{C}}\right]+\text{k}\text{z}\Longleftrightarrow$$ $$\frac{\text{a}-\text{b}}{\text{C}}=\frac{\text{x}-\text{y}}{\text{C}}\cdot e^{\text{k}\text{z}}\Longleftrightarrow$$ $$\text{a}-\text{b}=\left(\text{x}-\text{y}\right)\cdot e^{\text{k}\text{z}}\Longleftrightarrow$$ $$\frac{\text{a}-\text{b}}{\text{x}-\text{y}}=e^{\text{k}\text{z}}\Longleftrightarrow$$ $$\text{k}=\frac{1}{\text{z}}\cdot\ln\left[\frac{\text{a}-\text{b}}{\text{x}-\text{y}}\right]$$

So, in your example:

$$\text{k}=\frac{1}{\frac{29}{60}}\cdot\ln\left[\frac{28.8-24.5}{28.0-24.5}\right]=\frac{60}{29}\cdot\ln\left[\frac{4.3}{3.5}\right]\approx0.4259008018016872$$

Answer 2

You want to eliminate one of $k, t$ so from the first equation, we get $Ce^{-kt} = 4.3$, plugging this into the second equation, we get $$3.5 = Ce^{-kt}\cdot \exp\left({-\frac{29k}{60}}\right) = 4.3 \exp \left(-\frac{29k}{60}\right)$$

which you can then use to solve for $k$ easily.

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Note, we use the property that $e^{a+b} = e^a \cdot e^b$.

Answer 3

Hint:

If $k$ is a constant you can put $y=Ce^{-kt}$ and your equation becomes: $$ 0.8-y=\frac{y}{e^{kb}} \quad \mbox{with} \quad b=\frac{29}{60} $$ that you can easily solve, than back-substitute use logarithms.

Answer 4

your equation system can be written as $$\frac{28.8-24.5}{C}=e^{-kt}$$ $$\frac{28.0-24.5}{C}=e^{-kt}\cdot e^{-\frac{29}{60}k}$$ plugging (1) in (2) we obtain $$\frac{28.0-24.5}{C}=\frac{28.8-24.5}{C}\cdot e^{-\frac{29}{60}k}$$ Can you proceed?