\(\displaystyle i = I e^{-t/T}\)
From the chart, we have i = 0.0156 when t = 0.05
Substituting these two values into the exponential model gives us equation #1.
\(\displaystyle 0.0156 = I e^{-0.05/T}\)
Likewise, substituting i = 0.0212 when t = 0.1 gives us equation #2.
\(\displaystyle 0.0121 = I e^{-0.1/T}\)
Now, there are different methods to solve this system of two equations for I and T.
I began with these two steps on each equation:
(1) Divide both sides by I
(2) Take the natural logarithm of each side
These steps yield the following equations.
\(\displaystyle ln(0.0156/I) = \frac{-0.05}{T}\)
\(\displaystyle ln(0.0121/I) = \frac{-0.1}{T}\)
Next, I applied a property of logarithms to the lefthand side of each of these results.
\(\displaystyle ln(0.0156) - ln(I) = \frac{-0.05}{T}\)
\(\displaystyle ln(0.0121) - ln(I) = \frac{-0.1}{T}\)
Since the term ln(I) appears in each equation, I eliminated it from the system by subtraction. In other words, I subtracted the second equation above from the first:
\(\displaystyle [ln(0.0156) - ln(I)] - [ln(0.0121) - ln(I)] = \frac{-0.05}{T} - \frac{-0.1}{T}\)
\(\displaystyle ln(0.0156) - ln(0.0121) - ln(I) + ln(I) = \frac{-0.05 - (-0.1)}{T}\)
\(\displaystyle ln(0.0156) - ln(0.0121) = \frac{0.05}{T}\)
I used a scientific calculator to evaluate the lefthand side above. I then solved for T.
\(\displaystyle 0.2541 = \frac{0.05}{T}\)
\(\displaystyle T = \frac{0.05}{0.2541} = 0.1968\)
Knowing the value of T allowed me to calculate the value of I, using either equation #1 or equation #2.
Substituting 0.1968 for T into equation #2 yields:
\(\displaystyle 0.0121 = I e^{-0.1/0.1968}\)
\(\displaystyle 0.0121 = I e^{-0.5081}\)
Again, using a scientific calculator, I evaluated the power of e above. I then solved for I.
\(\displaystyle 0.0121 = 0.6016 I\)
\(\displaystyle I = \frac{0.0121}{0.6016} = 0.0201\)
From the given chart, I formed five pairs of equations (i.e., systems) from the given values of i and t.
As long as the results for I and T are rounded to two places, they always turn out to be 0.02 and 0.2, respectively.
This approach allows one to determine the answers in this exercise, but it has nothing to do with "linearizing using logs". Maybe that is associated with the regression method. What topic is your class currently covering?
Use the model \(\displaystyle i = 0.02 e^{-t/0.2}\) to answer parts (b) and (c).
Hmmm, I just noticed that part (b) asks for I instead of i, when t = 12. I'm not sure why. Maybe it's a typo?
Also, 12 milliseconds is so far outside the data in the chart that our values of I and T might not apply because of bad extrapolation. Maybe it's supposed to be 1.2? :?