Suppose you want to have $$ \$ 300,000 $$ for retirement in 20 years. Your account earns $$ 5 \% $$ interest. How much would you need to deposit in the account each month?

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We are given that we want an amount
$$
A = \$300,\!000
$$
at the end of $$t = 20$$ years, and the account earns an annual interest rate of $$5\%$$ compounded monthly.

Let  
$$
r = \frac{0.05}{12}
$$
be the monthly interest rate and  
$$
n = 20 \times 12 = 240
$$
be the number of monthly deposits.

Using the future value of an annuity formula for monthly deposits:
$$
A = P \frac{(1+r)^n - 1}{r},
$$
we need to solve for the monthly deposit $$P$$.

Rearrange the formula to solve for $$P$$:
$$
P = A \cdot \frac{r}{(1+r)^n - 1}.
$$

Substitute the known values:
$$
P = 300,\!000 \times \frac{\frac{0.05}{12}}{(1+\frac{0.05}{12})^{240} - 1}.
$$

Step 1. Compute the monthly interest rate:
$$
r = \frac{0.05}{12} \approx 0.00416667.
$$

Step 2. Compute the growth factor for 240 months:
$$
(1+r)^{240} \approx (1.00416667)^{240}.
$$
Taking logarithms, we have:
$$
\ln(1.00416667) \approx 0.00415888,
$$
so
$$
240 \times 0.00415888 \approx 0.99813,
$$
and
$$
(1.00416667)^{240} \approx e^{0.99813} \approx 2.714.
$$

Step 3. Compute the denominator:
$$
(1+r)^n - 1 \approx 2.714 - 1 = 1.714.
$$

Step 4. Now substitute into the formula for $$P$$:
$$
P \approx 300,\!000 \times \frac{0.00416667}{1.714}.
$$

Step 5. Calculate the fraction:
$$
\frac{0.00416667}{1.714} \approx 0.002430.
$$

Step 6. Finally, compute:
$$
P \approx 300,\!000 \times 0.002430 \approx 729.
$$

Thus, you need to deposit approximately $$\$729$$ each month.
You need to deposit approximately \$729 each month.

Suppose you want to have \( \$ 300,000 \) for retirement in 20 years. Your account earns \( 5 \% \) interest. How much would you need to deposit in the account each month?

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