$$ 1

Question

$$ 1<\quad \begin{array}{l}	ext { In } 1850 	ext {, a person sold a house to a lady for } \$ 30 	ext {. If the lady } \ 	ext { had put the } \$ 30 	ext { into a bank account paying } 6 \% 	ext { interest, how } \ 	ext { much would the investment have been worth in the year } \ 2010 	ext { if interest were compounded in the following ways? } \ 	ext { a. monthly } \ 	ext { b. continuously } \ 	ext { a. If compounded monthly, the investment would be worth } \ \$ \square 	ext { in } 2010 	ext {. } \ 	ext { (Round to the nearest dollar as needed.) }\end{array} $$

UpStudy AI · Accepted Answer

Let $$ P=30 $$ dollars be the principal, $$ r=0.06 $$ (6% annual interest), $$ n=12 $$ (monthly compounding), and $$ t=160 $$ years (since $$2010-1850=160$$).

1. The compound interest formula is  
$$
A = P\left(1+\frac{r}{n}\right)^{nt}.
$$

2. Substituting the values, we have  
$$
A = 30\left(1+\frac{0.06}{12}\right)^{12\cdot 160} = 30\left(1.005\right)^{1920}.
$$

3. To approximate $$ \left(1.005\right)^{1920} $$, note that  
$$
\left(1.005\right)^{1920} = e^{1920\ln(1.005)}.
$$
Since  
$$
\ln(1.005) \approx 0.0049875,
$$
we have  
$$
1920\ln(1.005) \approx 1920\times 0.0049875 \approx 9.576.
$$
Thus,  
$$
\left(1.005\right)^{1920} \approx e^{9.576} \approx 14413.
$$

4. Multiplying by 30, we obtain  
$$
A \approx 30\times 14413 \approx 432390.
$$

So, if the lady had invested \$30 in 1850 at 6% interest compounded monthly, by 2010 the investment would be worth approximately \$432,390 (rounded to the nearest dollar).
By investing \$30 in 1850 at 6% interest compounded monthly, the investment would be worth approximately \$432,390 in 2010.

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