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3300 dollars is placed in an account with an annual interest rate of
8.25%. To the nearest tenth of a year, how long will it take for the account value to reach 7500 dollars?
Answer:

$3300$ dollars is placed in an account with an annual interest rate of $8.25 \%$ . To the nearest tenth of a year, how long will it take for the account value to reach $7500$ dollars? $\newline$ Answer:

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Exponential growth and decay: word problems

Full solution

Q.

3300

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

8.25 \%

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

7500

dollars?

\newline

Answer:

Determine type of interest: Determine the type of interest being applied. $\newline$ Since the problem does not specify compound or simple interest, we will assume compound interest is applied annually.
Identify compound interest formula: Identify the formula for compound interest. $\newline$ The formula for compound interest is $A = P(1 + \frac{r}{n})^{nt}$ , where: $\newline$ $A$ = the amount of money accumulated after $n$ years, including interest. $\newline$ $P$ = the principal amount (the initial amount of money). $\newline$ $r$ = the annual interest rate (decimal). $\newline$ $n$ = the number of times that interest is compounded per year. $\newline$ $t$ = the time the money is invested for, in years.
Convert interest rate to decimal: Convert the annual interest rate from a percentage to a decimal. $8.25\%$ as a decimal is $0.0825$ .
Set up equation and solve: Since the interest is compounded annually, $n$ will be $1$ .
Simplify equation and solve: Set up the equation with the given values and solve for $t$ . $A = 7500, P = 3300, r = 0.0825, n = 1.$ $7500 = 3300(1 + 0.0825/1)^{(1\cdot t)}$
Take natural logarithm: Simplify the equation and solve for $t$ . $\newline$ $7500 = 3300(1 + 0.0825)^t$ $\newline$ $\frac{7500}{3300} = (1.0825)^t$ $\newline$ $2.2727 \approx (1.0825)^t$
Calculate natural logarithm: Take the natural logarithm of both sides to solve for $t$ . $\ln(2.2727) \approx t \times \ln(1.0825)$
Round answer: Calculate the natural logarithm of both sides. $\newline$ $\ln(2.2727) \approx t \times \ln(1.0825)$ $\newline$ $t \approx \frac{\ln(2.2727)}{\ln(1.0825)}$ $\newline$ $t \approx \frac{0.8195}{0.0799}$ $\newline$ $t \approx 10.25$
Round answer: Calculate the natural logarithm of both sides. $\newline$ $\ln(2.2727) \approx t \times \ln(1.0825)$ $\newline$ $t \approx \frac{\ln(2.2727)}{\ln(1.0825)}$ $\newline$ $t \approx \frac{0.8195}{0.0799}$ $\newline$ $t \approx 10.25$ Round the answer to the nearest tenth of a year. $\newline$ $t \approx 10.3$ years

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