A person invested $\$ 450$ in an account growing at a rate allowing the money to double every $6$ years. How much money would be in the account after $28$ years, to the nearest dollar? $\newline$ Answer:

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Exponential growth and decay: word problems

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Q. A person invested

\$ 450

in an account growing at a rate allowing the money to double every

6

years. How much money would be in the account after

28

years, to the nearest dollar?

\newline

Answer:

Identify growth type: Determine the type of growth. The account doubles every $6$ years, indicating exponential growth.
Find initial amount and factor: Identify the initial amount $a$ and the growth factor $b$ . The initial amount $a$ is $\$450$ , and since the money doubles, the growth factor $b$ is $2$ .
Calculate number of doubles: Calculate $x$ , the number of times the initial amount doubles. The time period $t$ is $28$ years, and the doubling period $T$ is $6$ years. $x = \frac{t}{T} = \frac{28}{6}$
Calculate amount after $24$ years: Since $28$ is not a multiple of $6$ , we need to find the closest multiple of $6$ that does not exceed $28$ . The closest multiple of $6$ to $28$ without going over is $24$ ( $6 \times 4$ ). So, $x = \frac{24}{6} = 4$ . This means the money will double $4$ times in $24$ years.
Determine growth rate per year: Use the exponential growth formula $P(x) = a(b)^x$ to calculate the amount after $24$ years. $\newline$ $P(x) = 450(2)^4$
Calculate amount after $28$ years: Evaluate the expression $450(2)^4$ . $\newline$ $450(2)^4 = 450 \times 16 = 7200$ $\newline$ After $24$ years, the account will have $\$7200$ .
Evaluate growth factor: Determine the growth over the remaining $4$ years. $\newline$ Since the account doubles every $6$ years, we need to find the growth rate per year. $\newline$ The growth rate per year is the $6$ th root of $2$ .
Final amount after $28$ years: Calculate the amount after the remaining $4$ years using the annual growth rate. $\newline$ Let $r$ be the annual growth rate, then $r = (2)^{\frac{1}{6}}$ . $\newline$ The amount after $28$ years will be $P(28) = 7200 \times r^4$ .
Final amount after $28$ years: Calculate the amount after the remaining $4$ years using the annual growth rate. $\newline$ Let $r$ be the annual growth rate, then $r = (2)^{1/6}$ . $\newline$ The amount after $28$ years will be $P(28) = 7200 \times r^4$ .Evaluate the expression 7200 \times (2)^{1/6}^4. $\newline$ First, calculate (2)^{1/6}^4 = (2)^{4/6} = (2)^{2/3}.
Final amount after $28$ years: Calculate the amount after the remaining $4$ years using the annual growth rate. $\newline$ Let $r$ be the annual growth rate, then $r = (2)^{1/6}$ . $\newline$ The amount after $28$ years will be $P(28) = 7200 \times r^4$ .Evaluate the expression 7200 \times (2)^{1/6}^4. $\newline$ First, calculate (2)^{1/6}^4 = (2)^{4/6} = (2)^{2/3}.Now, calculate $(2)^{2/3}$ . $\newline$ This is the cube root of $2$ squared, which is approximately $1.5874$ .
Final amount after $28$ years: Calculate the amount after the remaining $4$ years using the annual growth rate. $\newline$ Let $r$ be the annual growth rate, then $r = (2)^{1/6}$ . $\newline$ The amount after $28$ years will be $P(28) = 7200 \times r^4$ .Evaluate the expression 7200 \times (2)^{1/6}^4. $\newline$ First, calculate (2)^{1/6}^4 = (2)^{4/6} = (2)^{2/3}.Now, calculate $(2)^{2/3}$ . $\newline$ This is the cube root of $2$ squared, which is approximately $1.5874$ .Multiply the amount after $24$ years by the growth factor for the remaining $4$ years. $\newline$ $P(28) = 7200 \times 1.5874$
Final amount after $28$ years: Calculate the amount after the remaining $4$ years using the annual growth rate. $\newline$ Let $r$ be the annual growth rate, then $r = (2)^{1/6}$ . $\newline$ The amount after $28$ years will be $P(28) = 7200 \times r^4$ .Evaluate the expression 7200 \times (2)^{1/6}^4. $\newline$ First, calculate (2)^{1/6}^4 = (2)^{4/6} = (2)^{2/3}.Now, calculate $(2)^{2/3}$ . $\newline$ This is the cube root of $2$ squared, which is approximately $1.5874$ .Multiply the amount after $24$ years by the growth factor for the remaining $4$ years. $\newline$ $P(28) = 7200 \times 1.5874$ Evaluate the expression $7200 \times 1.5874$ . $\newline$ $P(28) \approx 7200 \times 1.5874 \approx 11429.28$ $\newline$ Round to the nearest dollar.
Final amount after $28$ years: Calculate the amount after the remaining $4$ years using the annual growth rate. $\newline$ Let $r$ be the annual growth rate, then $r = (2)^{\frac{1}{6}}$ . $\newline$ The amount after $28$ years will be $P(28) = 7200 \times r^4$ .Evaluate the expression 7200 \times (2)^{\frac{1}{6}}^4. $\newline$ First, calculate (2)^{\frac{1}{6}}^4 = (2)^{\frac{4}{6}} = (2)^{\frac{2}{3}}.Now, calculate $(2)^{\frac{2}{3}}$ . $\newline$ This is the cube root of $2$ squared, which is approximately $1.5874$ .Multiply the amount after $24$ years by the growth factor for the remaining $4$ years. $\newline$ $P(28) = 7200 \times 1.5874$ Evaluate the expression $7200 \times 1.5874$ . $\newline$ $P(28) \approx 7200 \times 1.5874 \approx 11429.28$ $\newline$ Round to the nearest dollar.The final amount in the account after $28$ years, rounded to the nearest dollar, is $r = (2)^{\frac{1}{6}}$ $0$ .