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We are given the problem:

- Initial investment ($ P_0 $): $5000
- Final amount ($ A $): $8200
- Interest rate ($ r $): 7.5% per year, compounded quarterly
- We need to find the time ($ t $) in years.

The formula for compound interest is:

\[
A = P_0 \left(1 + \frac{r}{n}\right)^{nt}
\]

Where:
- $ A $ is the final amount.
- $ P_0 $ is the initial principal (investment).
- $ r $ is the annual interest rate (decimal form).
- $ n $ is the number of times the interest is compounded per year.
- $ t $ is the number of years.

### Step 1: Plug in the known values
- $ A = 8200 $
- $ P_0 = 5000 $
- $ r = \frac{7.5}{100} = 0.075 $ (convert the percentage to decimal)
- $ n = 4 $ (since it's compounded quarterly)

We now plug these into the formula:

\[
8200 = 5000 \left(1 + \frac{0.075}{4}\right)^{4t}
\]

### Step 2: Simplify the equation
First, calculate $ 1 + \frac{0.075}{4} $:

\[
1 + \frac{0.075}{4} = 1 + 0.01875 = 1.01875
\]

So, the equation becomes:

\[
8200 = 5000 \times (1.01875)^{4t}
\]

### Step 3: Isolate the exponential term
Divide both sides by 5000:

\[
\frac{8200}{5000} = (1.01875)^{4t}
\]

\[
1.64 = (1.01875)^{4t}
\]

### Step 4: Apply logarithms to solve for $ t $
Take the natural logarithm of both sides:

\[
\ln(1.64) = \ln\left((1.01875)^{4t}\right)
\]

Using the property of logarithms $ \ln(a^b) = b\ln(a) $, we get:

\[
\ln(1.64) = 4t \ln(1.01875)
\]

Now solve for $ t $:

\[
t = \frac{\ln(1.64)}{4 \ln(1.01875)}
\]

### Step 5: Calculate the result

Using approximate logarithm values:

\[
\ln(1.64) \approx 0.4947, \quad \ln(1.01875) \approx 0.01856
\]

Substitute into the equation:

\[
t = \frac{0.4947}{4 \times 0.01856} = \frac{0.4947}{0.07424} \approx 6.66 \text{ years}
\]

### Final Answer:
The time required for the investment to grow to $8200 is approximately **6.66 years**.

Let me know if you would like more details or further clarification.

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Here are 5 related questions to consider:
1. How does the compounding frequency affect the growth of investments?
2. What would the time be if the interest were compounded annually instead of quarterly?
3. How would a different interest rate (e.g., 5% or 10%) change the time needed for growth?
4. What is the relationship between the initial investment and the final amount?
5. Can we derive a general formula for finding the time $ t $ based on different interest rates and compounding frequencies?

**Tip:** Always check whether the interest rate is compounded yearly, quarterly, or monthly, as this can significantly affect the time required for an investment to grow.

Find the time required for an investment of 5000 dollars to grow to 8200 dollars at an interest rate of 7.5 percent per year, compounded quarterly.

Learn how to calculate the time required for a $5000 investment to grow to $8200 at an annual interest rate of 7.5%, compounded quarterly. This problem uses the compound interest formula and logarithmic properties for solving. Suitable for high school students.

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