A person invested dollar sign, 3, comma, 900$3,900 in an account growing at a rate allowing the money to double every 7 years. How much money would be in the account after 18 years, *to the nearest dollar*?

The key idea is that the investment doubles every 7 years. Let's break this problem down step by step.

### Step 1: Determine the number of doubling periods in 18 years.
Since the money doubles every 7 years, we divide the total time by 7 to see how many doubling periods there are:

\[
\frac{18}{7} \approx 2.57
\]

This means we have slightly over 2 full doubling periods in 18 years.

### Step 2: Calculate the money after each doubling period.

- Initially, the person invested $3,900.
- After the first 7 years, the investment doubles: 
  \[
  3,900 \times 2 = 7,800
  \]
- After the next 7 years (14 years in total), it doubles again:
  \[
  7,800 \times 2 = 15,600
  \]
  
Since we have 2.57 doubling periods, we are past two full doubling periods but not yet at three.

### Step 3: Use the formula for exponential growth.
To handle fractional doubling, use the formula for exponential growth:

\[
A = P \times 2^{t/T}
\]

Where:
- $ A $ is the amount of money after time $ t $,
- $ P $ is the initial investment ($3,900),
- $ t $ is the time period (18 years),
- $ T $ is the doubling time (7 years).

Now, plug in the values:

\[
A = 3,900 \times 2^{18/7}
\]

Let's calculate this:

\[
A = 3,900 \times 2^{2.57} \approx 3,900 \times 5.997
\]

\[
A \approx 23,388
\]

### Final Answer:
The amount of money in the account after 18 years is approximately **$23,388**.

Would you like more details or have any questions?

Here are some related questions to explore:
1. How would the outcome change if the doubling period were 5 years instead of 7?
2. What would happen if the interest rate changed every year?
3. How much money would be there after 21 years?
4. Can we use logarithms to reverse-engineer the time required for a specific target amount?
5. What if the investment was compounded continuously instead of doubling every fixed period?

**Tip:** Exponential growth processes (like this one) are common in finance, biology, and physics—recognizing them helps solve a wide range of problems quickly.

A person invested $3,900 in an account growing at a rate allowing the money to double every 7 years. How much money would be in the account after 18 years, to the nearest dollar?

Find out how $3,900 grows over 18 years with exponential growth. This example explains how money doubles every 7 years using the formula A = P × 2^(t/T). Calculated amount after 18 years is approximately $23,388. Suitable for Grades 9-12.

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