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4900 dollars is placed in an account with an annual interest rate of
5.25%. To the nearest tenth of a year, how long will it take for the account value to reach 13200 dollars?
Answer:

$4900$ dollars is placed in an account with an annual interest rate of $5.25 \%$ . To the nearest tenth of a year, how long will it take for the account value to reach $13200$ dollars? $\newline$ Answer:

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Exponential growth and decay: word problems

Full solution

Q.

4900

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

5.25 \%

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

13200

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 $n = 1$ .
Set Up Equation: Set up the equation with the given values. $\newline$ We need to find $t$ when $A = 13200$ , $P = 4900$ , $r = 5.25\%$ (or $0.0525$ as a decimal), and $n = 1$ . $\newline$ $13200 = 4900(1 + 0.0525/1)^{(1\cdot t)}$
Simplify Equation: Simplify the equation. $\newline$ $13200 = 4900(1 + 0.0525)^{t}$ $\newline$ $13200 = 4900(1.0525)^{t}$
Isolate Exponential Part: Divide both sides by $4900$ to isolate the exponential part of the equation. $\newline$ $\frac{13200}{4900} = (1.0525)^{(t)}$ $\newline$ $2.69387755102 \approx (1.0525)^{(t)}$
Take Natural Logarithm: Take the natural logarithm of both sides to solve for $t$ . $\ln(2.69387755102) = \ln((1.0525)^{t})$ $\ln(2.69387755102) = t \cdot \ln(1.0525)$
Calculate Value of t: Calculate the value of $t$ . $t = \frac{\ln(2.69387755102)}{\ln(1.0525)}$ $t \approx 20.1094$
Round Answer: Round the answer to the nearest tenth of a year. $t \approx 20.1$ years

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