Suppose that $\$ 15,000$ is invested and at the end of $3 \mathrm{yr}$ , the value of the account is $\$ 19,356.92$ . Use the model $A=P e^{r t}$ to determine the average rate of return $r$ under continuous compounding.

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Grade 8

Compound interest

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Q. Suppose that

\$ 15,000

is invested and at the end of

3 \mathrm{yr}

, the value of the account is

\$ 19,356.92

. Use the model

A=P e^{r t}

to determine the average rate of return

r

under continuous compounding.

Write Down Given Information: First, let's write down what we know: $\newline$ Initial investment $P$ = $\$15,000$ $\newline$ Final amount $A$ = $\$19,356.92$ $\newline$ Time $t$ = $3$ years $\newline$ We need to find the rate $r$ .
Use Continuous Compounding Formula: We use the formula for continuous compounding: $A = Pe^{rt}$ . Let's plug in the values we know: $\$19,356.92 = \$15,000 \times e^{r\times 3}$ .
Solve for Rate: Now, we need to solve for $r$ . Let's start by dividing both sides by $\$15,000$ . $\$19,356.92 / \$15,000 = e^{r*3}$
Calculate Natural Logarithm: Calculate the left side: $19356.92 / 15000 = 1.290461333$ . $\newline$ So, $1.290461333 = e^{r*3}$ .
Apply Natural Logarithm Property: Next, we take the natural logarithm ( $\ln$ ) of both sides to get rid of the exponential. $\ln(1.290461333) = \ln(e^{r\cdot 3})$
Calculate Natural Logarithm: The natural logarithm of $e$ to the power of something is just that something, so: $\newline$ $\ln(1.290461333) = r \times 3$
Divide to Solve for Rate: Calculate the natural logarithm of $1.290461333$ : $\ln(1.290461333) \approx 0.2552725051$ . So, $0.2552725051 = r \times 3$ .
Calculate Average Rate of Return: Now, divide both sides by $3$ to solve for $r$ . $\frac{0.2552725051}{3} = r$
Calculate Average Rate of Return: Now, divide both sides by $3$ to solve for $r$ . $\newline$ $\frac{0.2552725051}{3} = r$ Calculate $r$ : $0.2552725051 / 3 \approx 0.08509083503$ . $\newline$ So, $r \approx 0.08509083503$ , which is the average rate of return.