To determine the amount to deposit now (the present value, $P$), we use the formula for compound interest:

$A = P \cdot \left(1 + \frac{r}{n}\right)^{n \cdot t}$

Where:

• $A = 5000$ (the future value),
• $r = 0.0684$ (the annual interest rate as a decimal),
• $n = 365$ (compounded daily),
• $t = 19$ years (time).

We rearrange the formula to solve for $P$:

$P = \frac{A}{\left(1 + \frac{r}{n}\right)^{n \cdot t}}$

Step 1: Plug in the known values

$P = \frac{5000}{\left(1 + \frac{0.0684}{365}\right)^{365 \cdot 19}}$

Step 2: Calculate $\frac{r}{n}$

$\frac{0.0684}{365} \approx 0.000187397$

$1 + \frac{0.0684}{365} \approx 1.000187397$

Step 3: Calculate $n \cdot t$

$365 \cdot 19 = 6935$

Step 4: Raise the base to the power $n \cdot t$

$(1.000187397)^{6935} \approx 4.017663$

Step 5: Calculate $P$

$P = \frac{5000}{4.017663} \approx 1244.75$

Final Answer:

You would need to deposit approximately $1,244.75 today to have $5,000 in the account in 19 years with 6.84% interest compounded daily.

Would you like to see a detailed breakdown of each calculation or explore similar problems?

Here are 5 related questions:

1. What if the compounding frequency were monthly instead of daily?
2. How does the required deposit change with a lower interest rate, such as 5%?
3. How much interest is earned over the 19 years in this scenario?
4. How would the calculations differ if the interest were compounded continuously?
5. What if you wanted $10,000 instead of $5,000 in 19 years?

Tip: Compounding frequency significantly affects the amount needed for future savings—more frequent compounding results in greater accumulated growth.