$3$ ) The amount of money $A$ after $t$ years that principle $P$ will become if it is invested at rate $r$ compounded $n$ times a year is given by the relationship $\mathrm{A}(t)=P\left(1+\frac{r}{n}\right)^{n t}$ where $r$ is expressed as a decimal. To the nearest tenth, how long will it take $\$ 2,500$ to become $\$ 4,500$ if it is invested at $7 \%$ and is compounded quarterly? [Show all work.]

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Compound interest: word problems

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3

) The amount of money

A

after

t

years that principle

P

will become if it is invested at rate

r

compounded

n

times a year is given by the relationship

\mathrm{A}(t)=P\left(1+\frac{r}{n}\right)^{n t}

where

r

is expressed as a decimal. To the nearest tenth, how long will it take

\$ 2,500

to become

\$ 4,500

if it is invested at

7 \%

and is compounded quarterly? [Show all work.]

Given Formula Identification: We are given the formula for compound interest: $A(t) = P(1 + \frac{r}{n})^{nt}$ . We need to find the time $t$ it takes for an initial investment $P$ of $\$2,500$ to grow to $A(t)$ of $\$4,500$ at an annual interest rate $r$ of $7\%$ compounded quarterly ( $n = 4$ times a year).
Values Identification: First, let's identify the values from the problem: $\newline$ $P = \$2,500$ $\newline$ $A(t) = \$4,500$ $\newline$ $r = 7\% = 0.07$ (as a decimal) $\newline$ $n = 4$ (compounded quarterly) $\newline$ We want to find $t$ .
Plug Values into Formula: Now, let's plug these values into the compound interest formula: $\newline$ $\$4,500 = \$2,500(1 + \frac{0.07}{4})^{4t}$
Isolate Term with t: Next, we need to isolate the term with $t$ on one side of the equation. To do this, we divide both sides by $\$2,500$ : $\$4,500 / \$2,500 = (1 + 0.07/4)^{4t}$
Take Natural Logarithm: Perform the division on the left side of the equation: $\newline$ $1.8 = (1 + 0.07/4)^{4t}$
Rewrite Right Side: Now, we need to solve for $t$ . First, we take the natural logarithm ( $\ln$ ) of both sides to remove the exponent on the right side: $\newline$ $\ln(1.8) = \ln((1 + 0.07/4)^{(4t)})$
Solve for t: Using the property of logarithms that $\ln(a^b) = b\cdot\ln(a)$ , we can rewrite the right side of the equation: $\newline$ $\ln(1.8) = 4t \cdot \ln(1 + \frac{0.07}{4})$
Calculate t: Now, we solve for $t$ by dividing both sides by $4\cdot\ln(1 + 0.07/4)$ : $t = \frac{\ln(1.8)}{4 \cdot \ln(1 + 0.07/4)}$
Final Answer: Let's calculate the value of $t$ using a calculator: $\newline$ $t \approx \frac{\ln(1.8)}{4 \times \ln(1 + \frac{0.07}{4})}$ $\newline$ $t \approx \frac{\ln(1.8)}{4 \times \ln(1.0175)}$ $\newline$ $t \approx \frac{0.587787}{4 \times 0.017328}$ $\newline$ $t \approx \frac{0.587787}{0.069312}$ $\newline$ $t \approx 8.481$
Final Answer: Let's calculate the value of $t$ using a calculator: $\newline$ $t \approx \frac{\ln(1.8)}{4 \times \ln(1 + \frac{0.07}{4})}$ $\newline$ $t \approx \frac{\ln(1.8)}{4 \times \ln(1.0175)}$ $\newline$ $t \approx \frac{0.587787}{4 \times 0.017328}$ $\newline$ $t \approx \frac{0.587787}{0.069312}$ $\newline$ $t \approx 8.481$ To the nearest tenth, the time $t$ it will take for the investment to grow from $\$2,500$ to $\$4,500$ is approximately $8$ . $5$ years.