Simple logarithmic equation 
If $y=Ae^{-kt}$ and $y=19.6$ when $t=2$, and $y=19.02$ when $t=5$, find the value of the constants $A$ and $k$. Give your answers correct to $2$ decimal places.

I have spent a while (an hour+) on this question, but I'm not sure how to solve it algebraically as there are two variables to find. 
 A: The first statement says that if you plug in $t=2$, you get $19.6$. That is,
$$ 19.6 = Ae^{-2k}.$$
The second statement says that if you plug in $t=5$, you get $19.02$, so
$$ 19.02 = Ae^{-5k}.$$
Now, solving for $A$ in each of the two equations we have:
$$A = 19.6e^{2k}\qquad\text{and}\qquad 19.02e^{5k} = A.$$
Since they are both equal to $A$, that means that they are equal to each other, so
$$19.6 e^{2k} = 19.02e^{5k}.$$
Dividing through by $e^{2k}$ and by $19.02$, we get
$$\frac{19.6}{19.02} = \frac{e^{5k}}{e^{2k}} = e^{3k}.$$
Now you can solve for $k$ using logarithms. Then use the equations for $A$ to solve for $A$ as well.
A: You have:
$$19.6 = A e^{-2k}$$
and
$$19.02 = A e^{-5k}$$
The above is a system of two equations in two unknowns. Surely, you can take it from here to find  $A$ and $k$.
If you are not allowed the use of calculators then you can use the following strategy:
Divide the two equations to get:
$$e^{3k} = \frac{19.6}{19.02}$$
Now, $\frac{19.6}{19.02} \approx 1.03$ and therefore $k$  has to be less than $\frac{1}{3}$ as $e > 2$. Therefore, we can approximate $e^{3k}$ using Taylor series upto two decimals by:
$$e^{3k} \approx 1 + 3k + \frac{(3k)^2}{2}$$
Thus, we have:
$$1 + 3k + \frac{(3k)^2}{2} = 1.03$$
Solve the quadratic to identify $k$ (be careful to rule out any spurious solutions for $k$). You can then get to $A$ using the same approximation.
In fact, solving the above quadratic for $k$ yields $0.009854$ and $-0.676$ as the two possible solutions. Note that $k$ cannot be less than $0$ for the original equation as $e^{3k}$ is increasing in $k$ and $e^0 < \frac{19.6}{19.02}$. Therefore, $k=0.009854$. The accurate solution for $k$ is given by $\frac{1}{3} \mathrm{log}(\frac{19.6}{19.02})=0.0100$.
