In 1850, a person sold a house to a lady for . If the lady had put the into a bank account paying interest, how much would the investment have been worth in the year 2010 if interest were compounded in the following ways?
a. monthly
b. continuously
a. If compounded monthly, the investment would be worth in 2010.
(Round to the nearest dollar as needed.)

Ask by Pollard Wells. in the United States

Mar 23,2025

Question

In 1850, a person sold a house to a lady for . If the lady had put the into a bank account paying interest, how much would the investment have been worth in the year 2010 if interest were compounded in the following ways?
a. monthly
b. continuously
a. If compounded monthly, the investment would be worth in 2010.
(Round to the nearest dollar as needed.)

Ask by Pollard Wells. in the United States

Mar 23,2025

UpStudy AI · Accepted Answer

Let

$$
P = 30, \quad r = 0.06, \quad n = 12, \quad t = 2010-1850 = 160.
$$

The compound interest formula is

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

Substituting the values for monthly compounding gives

$$
A = 30\left(1 + \frac{0.06}{12}\right)^{12 \times 160} = 30\left(1.005\right)^{1920}.
$$

Taking the natural logarithm to simplify the exponent,

$$
\ln\left((1.005)^{1920}\right) = 1920 \cdot \ln(1.005),
$$

and since

$$
\ln(1.005) \approx 0.0049875,
$$

we have

$$
1920 \cdot 0.0049875 \approx 9.576.
$$

Thus,

$$
A \approx 30 \cdot e^{9.576} \approx 30 \cdot 14418 \approx 432540.
$$

Rounded to the nearest dollar, if compounded monthly, the investment would be worth

$$
\boxed{\$432,540}
$$ in 2010.
If compounded monthly, the investment would be worth approximately \$432,540 in 2010.