To find the exponential function of the form $A(t) = Ie^{rt}$ that passes through the points $(4, 120)$ and $(10, 35)$, we can use logarithms as follows:

1. Set up the equations:

   The general form of the exponential function is: $A(t) = Ie^{rt}$

   For the point $(4, 120)$: $120 = Ie^{4r}$

   For the point $(10, 35)$: $35 = Ie^{10r}$

2. Solve the equations using logarithms:

   To eliminate $I$, we can divide the second equation by the first equation: $\frac{35}{120} = \frac{Ie^{10r}}{Ie^{4r}}$

   Simplify the equation: $\frac{35}{120} = e^{10r - 4r}$

   $\frac{35}{120} = e^{6r}$

   Simplify further: $\frac{35}{120} = \frac{7}{24}$

   $\frac{7}{24} = e^{6r}$

3. Solve for $r$:

   Take the natural logarithm of both sides: $\ln\left(\frac{7}{24}\right) = \ln(e^{6r})$

   $\ln\left(\frac{7}{24}\right) = 6r$

   $r = \frac{\ln\left(\frac{7}{24}\right)}{6}$

   Calculate the value: $r \approx \frac{\ln(7) - \ln(24)}{6}$

   $r \approx \frac{1.9459 - 3.1781}{6}$

   $r \approx \frac{-1.2322}{6}$

   $r \approx -0.20537$

4. Find $I$:

   Use the value of $r$ in one of the original equations to find $I$. Let's use the first equation $(4, 120)$: $120 = Ie^{4 \cdot -0.20537}$

   $120 = Ie^{-0.82148}$

   $120 = I \cdot 0.43991$

   $I \approx \frac{120}{0.43991}$

   $I \approx 272.76$

So, the exponential function is: $A(t) = 272.76e^{-0.20537t}$

Would you like more details or have any questions? Here are some further questions you might consider:

1. How can you confirm that the function fits the given points?
2. What does a negative $r$ indicate about the behavior of the function?
3. How can you apply this method to more than two points?
4. How does the initial value $I$ affect the graph of the function?
5. What are some real-life applications of exponential decay functions?

Tip: When solving for $r$ using logarithms, ensure you apply properties of logarithms correctly to simplify the process.