# 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.

-

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.

-
$k=0.0100128$ $A=19.9965$ –  dikuve Dec 17 '11 at 5:38
@dikuve: And the reason you believe that just giving out the solution rather than let the OP work it out for him or herself is... –  Arturo Magidin Dec 17 '11 at 5:44
actually you have solved the problem I am just providing the end result...He/she was asking about the procedure to solve it algebraically. So if you wish that he /she should work on it then you should start like this.. Let you have $y_1$ for $t_1$ and $y_2$ for $t_2$ using this you obtain two equation and two unknown for example $y_1=Ae^{-kt_1}$ and $y_2=Ae^{-kt_2}$ using these two equations you can find $A=y_1e^{kt_1}=y_2e^{kt_2}$ using this you can obtain $k$ after this either using $A=y_1e^{kt_1}$ or using $A=y_2e^{kt_2}$ you can obtain $A$. –  dikuve Dec 17 '11 at 6:18
@dikuve: If you think there is a better way of addressing the poster's question, then post it as an answer instead of telling me how I "should" have written mine. It's not as if I couldn't have found the final answer myself, so perhaps you could have stopped and wonder why I didn't include it instead. –  Arturo Magidin Dec 17 '11 at 6:19
Thanks for your valuable comment. In future I will keep in mind. –  dikuve Dec 17 '11 at 6:26

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 $\frac{1}{3} \mathrm{log}(\frac{19.6}{19.02})=0.0100$.

-
I got to that, but I'm not sure where to go from this stage. I might divide it but I am not permitted a calculator. –  tina nyaa Dec 17 '11 at 5:12
@tina nyaa: You could find explicit expressions for $k$ and $A$ without a calculator. However, finding $2$ decimal place answers without a calculator is not practical. (In the old days, you could instead have used tables or a slide rule. And now you could use fancy estimates for the logarithm, but I doubt that is intended.) –  André Nicolas Dec 17 '11 at 5:35