4. An initial amount of $$ \$ 2300 $$ is deposited in an account. In 5 years, the investment grows to $$ \$ 4237.61 $$. a. Find a formula for the value of the investment as a function of time, assuming exponential growth. Report any calculated values correct to two decimal places. Your formula should begin with " $$ y= $$ ". b. What annual interest rate does the account pay?

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To solve this problem, we will follow these steps:

### Part a: Find the formula for the value of the investment

1. **Identify the known values:**
   - Initial amount (principal) $$ P = 2300 $$
   - Amount after 5 years $$ A = 4237.61 $$
   - Time $$ t = 5 $$ years

2. **Use the formula for exponential growth:**
   The formula for exponential growth is given by:
   $$
   A = P e^{rt}
   $$
   where:
   - $$ A $$ is the amount of money accumulated after time $$ t $$,
   - $$ P $$ is the principal amount (initial investment),
   - $$ r $$ is the annual interest rate (as a decimal),
   - $$ t $$ is the time in years,
   - $$ e $$ is the base of the natural logarithm.

3. **Substituting the known values into the formula:**
   $$
   4237.61 = 2300 e^{5r}
   $$

4. **Solve for $$ e^{5r} $$:**
   $$
   e^{5r} = \frac{4237.61}{2300}
   $$

5. **Calculate $$ \frac{4237.61}{2300} $$:**
   Let's calculate this value.

Now, I will perform the calculation for $$ \frac{4237.61}{2300} $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{4237.61}{2300}$$
- step1: Convert the expressions:
  $$\frac{\frac{423761}{100}}{2300}$$
- step2: Multiply by the reciprocal:
  $$\frac{423761}{100}\times \frac{1}{2300}$$
- step3: Multiply the fractions:
  $$\frac{423761}{100\times 2300}$$
- step4: Multiply:
  $$\frac{423761}{230000}$$
The calculation gives us:
$$
e^{5r} \approx 1.842439
$$

6. **Take the natural logarithm of both sides:**
   $$
   5r = \ln(1.842439)
   $$

7. **Calculate $$ \ln(1.842439) $$:**
   Now, I will calculate the natural logarithm of $$ 1.842439 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\ln{\left(1.842439\right)}$$
- step1: Convert the expressions:
  $$\ln{\left(\frac{1842439}{1000000}\right)}$$
- step2: Rewrite the expression:
  $$\ln{\left(1842439\right)}-6\ln{\left(10\right)}$$
The calculation gives us:
$$
5r = \ln(1.842439) \approx 0.620576
$$

8. **Solve for $$ r $$:**
   $$
   r = \frac{0.620576}{5} \approx 0.1241152
   $$

9. **Convert $$ r $$ to a percentage:**
   $$
   r \approx 0.1241152 \times 100 \approx 12.41\%
   $$

10. **Write the formula for the value of the investment:**
   Now we can express the value of the investment as a function of time $$ t $$:
   $$
   y = 2300 e^{0.1241152t}
   $$

### Part b: What annual interest rate does the account pay?

The annual interest rate is approximately $$ 12.41\% $$.

### Final Answers:
a. The formula for the value of the investment is:
$$
y = 2300 e^{0.1241152t}
$$

b. The annual interest rate is approximately $$ 12.41\% $$.
a. The value of the investment after time $$ t $$ is $$ y = 2300 e^{0.1241t} $$. b. The annual interest rate is approximately $$ 12.41\% $$.

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