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4500 dollars is placed in an account with an annual interest rate of
7%. To the nearest tenth of a year, how long will it take for the account value to reach 12800 dollars?
Answer:

$4500$ dollars is placed in an account with an annual interest rate of $7 \%$ . To the nearest tenth of a year, how long will it take for the account value to reach $12800$ dollars? $\newline$ Answer:

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Exponential growth and decay: word problems

Full solution

Q.

4500

dollars is placed in an account with an annual interest rate of

7 \%

. To the nearest tenth of a year, how long will it take for the account value to reach

12800

dollars?

\newline

Answer:

Identify Formula: Identify the formula to use for compound interest. $\newline$ The formula for compound interest is $A = P(1 + r/n)^{(nt)}$ , where: $\newline$ $A$ is the amount of money accumulated after $n$ years, including interest. $\newline$ $P$ is the principal amount (the initial amount of money). $\newline$ $r$ is the annual interest rate (decimal). $\newline$ $n$ is the number of times that interest is compounded per year. $\newline$ $t$ is the time the money is invested for, in years. $\newline$ Since the problem does not specify how often the interest is compounded, we will assume it is compounded annually, so $n = 1$ .
Convert Rate to Decimal: Convert the annual interest rate from a percentage to a decimal. The annual interest rate is $7\%$ , which as a decimal is $0.07$ .
Set Up Equation: Set up the equation with the given values and solve for $t$ . We have $P = 4500$ , $A = 12800$ , $r = 0.07$ , and $n = 1$ . We need to find $t$ . $12800 = 4500(1 + 0.07/1)^{(1\cdot t)}$
Simplify and Solve: Simplify the equation and solve for $t$ . $12800 = 4500(1 + 0.07)^t$ $12800 = 4500(1.07)^t$ Now, divide both sides by $4500$ to isolate the exponential part. $\frac{12800}{4500} = (1.07)^t$ $2.84 \approx (1.07)^t$
Use Logarithms: Use logarithms to solve for $t$ . To solve for $t$ , we take the natural logarithm ( $\ln$ ) of both sides. $\ln(2.84) = \ln((1.07)^t)$ Now, use the property of logarithms that $\ln(a^b) = b\cdot\ln(a)$ . $\ln(2.84) = t \cdot \ln(1.07)$
Isolate and Calculate: Isolate $t$ and calculate its value. $t = \frac{\ln(2.84)}{\ln(1.07)}$ $t \approx 17.6725$
Round to Nearest Tenth: Round the answer to the nearest tenth of a year. $\newline$ $t \approx 17.7$ years

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