In​ 2004, an art collector paid ​$108 comma 467 comma 000 for a particular painting. The same painting sold for ​$28 comma 000 in 1950. Complete parts​ (a) through​ (d). Question content area bottom Part 1 ​a) Find the exponential growth rate​ k, to three decimal​ places, and determine the exponential growth function​ V, for which​ V(t) is the​ painting's value, in​ dollars, t years after 1950.

We are given two values for the painting: 
- In 1950, the value of the painting was $ V(0) = 28,000 $ dollars.
- In 2004 (54 years later), the value of the painting was $ V(54) = 108,467,000 $ dollars.

We are asked to find the exponential growth rate $ k $ and the exponential growth function $ V(t) $, where $ t $ represents the number of years after 1950.

The general form of an exponential growth function is:
\[
V(t) = V_0 e^{kt}
\]
where:
- $ V(t) $ is the value of the painting at time $ t $,
- $ V_0 $ is the initial value of the painting (in 1950),
- $ k $ is the growth rate,
- $ t $ is the number of years since 1950.

### Step 1: Set up the equation
We know:
- $ V(0) = 28,000 $, so $ V_0 = 28,000 $,
- $ V(54) = 108,467,000 $.

Thus, the equation for $ V(54) $ becomes:
\[
108,467,000 = 28,000 e^{54k}
\]

### Step 2: Solve for $ k $
Divide both sides of the equation by 28,000 to isolate the exponential term:
\[
\frac{108,467,000}{28,000} = e^{54k}
\]
\[
3873.107 = e^{54k}
\]

Take the natural logarithm (ln) of both sides to solve for $ k $:
\[
\ln(3873.107) = 54k
\]
\[
\ln(3873.107) \approx 8.261
\]
\[
8.261 = 54k
\]
\[
k = \frac{8.261}{54} \approx 0.153
\]

Thus, the exponential growth rate $ k $ is approximately $ 0.153 $ (to three decimal places).

### Step 3: Write the exponential growth function
Now that we know $ k $, the exponential growth function becomes:
\[
V(t) = 28,000 e^{0.153t}
\]

This is the exponential growth function for the painting's value.

---

Would you like more details or have any questions about these steps?

### Follow-up questions:
1. How does changing the growth rate $ k $ affect the exponential growth function?
2. What would be the value of the painting in 1970 using the exponential model?
3. How would the function change if the painting depreciated over time instead of growing?
4. Can you explain the difference between exponential growth and linear growth?
5. How could the exponential growth model be applied to other forms of investment?

### Tip:
When solving for the growth rate $ k $ in exponential equations, always remember to take the natural logarithm (ln) when isolating the exponent.

In 2004, an art collector paid $108,467,000 for a particular painting. The same painting sold for $28,000 in 1950. Complete parts (a) through (d). Part 1 (a) Find the exponential growth rate k, to three decimal places, and determine the exponential growth function V, for which V(t) is the painting's value, in dollars, t years after 1950.

Solve for the exponential growth rate and function for a painting that grew in value from $28,000 in 1950 to $108,467,000 in 2004. This problem covers exponential growth concepts, logarithmic equations, and algebraic methods, suitable for Grades 11-12 and college-level students.

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