Question

suppose you start saving today for a $70,000 down payment that you plan to make on a house in 7 years. assume that you make no deposits into the account after the initial deposit. for the account described below, how much would you have to deposit now to reach your $70,000 goal in 7 years
an account with daily compounding and an apr of 8%

you should invest $
(do not round until the final answer. then round to two decimal places as needed.)

Explanation:

Step1: Recall the compound - interest formula for present value

The formula for compound interest is $A = P(1+\frac{r}{n})^{nt}$, where $A$ is the future value, $P$ is the principal (present value), $r$ is the annual percentage rate (APR) (in decimal), $n$ is the number of times interest is compounded per year, and $t$ is the number of years. We want to solve for $P$, so we can re - arrange the formula to $P=\frac{A}{(1 + \frac{r}{n})^{nt}}$.

Step2: Identify the values of $A$, $r$, $n$, and $t$

• We know that $A=\$70000$ (the future value we want to achieve).
• The APR $r = 8\%=0.08$.
• Since the interest is compounded daily, $n = 365$ (assuming a non - leap year; if we consider leap years, in the long run, using 365 is a common approximation).
• The time $t = 7$ years.

Step3: Substitute the values into the formula

First, calculate the exponent $nt$: $nt=365\times7 = 2555$.
Then, calculate $\frac{r}{n}=\frac{0.08}{365}\approx0.000219178$.
Next, calculate $1+\frac{r}{n}=1 + 0.000219178=1.000219178$.
Then, calculate $(1+\frac{r}{n})^{nt}=(1.000219178)^{2555}$. We can use a calculator to find that $(1.000219178)^{2555}\approx e^{0.08\times7}$ (using the approximation $(1+\frac{r}{n})^{nt}\approx e^{rt}$ for large $n$, but we can also calculate it directly). Using a calculator, $(1.000219178)^{2555}\approx1.75067$.
Now, $P=\frac{70000}{(1.000219178)^{2555}}=\frac{70000}{1.75067}\approx39983.54$. (We can also calculate it more precisely:
$(1+\frac{0.08}{365})^{365\times7}=(1 + \frac{0.08}{365})^{2555}$. Let's compute this step - by - step.
First, $\frac{0.08}{365}\approx0.0002191780822$.
Then $1+\frac{0.08}{365}\approx1.0002191780822$.
Raise this to the power of 2555: $y = 1.0002191780822^{2555}$. Take the natural logarithm: $\ln(y)=2555\times\ln(1.0002191780822)\approx2555\times0.00021915\approx0.5599$. Then $y = e^{0.5599}\approx1.7505$. Then $P=\frac{70000}{1.7505}\approx39987.43$ (more precise calculation using calculator for the exponent:
$\ln(1.0002191780822)\approx2.1915\times10^{-4}$
$2555\times2.1915\times10^{-4}=2555\times0.00021915 = 0.5599$
$e^{0.5599}\approx1.7505$
$P=\frac{70000}{1.7505}\approx39987.43$ (if we use a more accurate value of $\ln(1.0002191780822)$: $\ln(1 + x)\approx x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\cdots$ for small $x$. Here $x = 0.0002191780822$, so $\ln(1 + x)\approx0.0002191780822-\frac{(0.0002191780822)^{2}}{2}+\frac{(0.0002191780822)^{3}}{3}\approx0.0002191780822-2.40\times10^{-8}+3.4\times10^{-12}\approx0.000219154$
Then $2555\times0.000219154 = 0.5599$
$e^{0.5599}\approx1.7505$
$P=\frac{70000}{1.7505}\approx39987.43$)

Answer:

$\$39987.43$ (The answer may vary slightly depending on the precision of the calculator used for the exponentiation. If we use a calculator to compute $(1+\frac{0.08}{365})^{365\times7}$ directly:
$(1+\frac{0.08}{365})^{2555}\approx(1.000219178)^{2555}\approx1.75067$
$P = \frac{70000}{1.75067}\approx39983.54$. But a more accurate calculation using a financial calculator or a calculator with high - precision exponentiation will give approximately $\$39987.43$)