Question $5$ : $\$ 500$ is invested in an account that pays $4.5 \%$ per annum, interest compounded monthly. Find how long it takes to reach $\$ 5000$ .

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Q. Question

5

\$ 500

is invested in an account that pays

4.5 \%

per annum, interest compounded monthly. Find how long it takes to reach

\$ 5000

Identify Variables: Identify the variables for the compound interest formula, which is $A = P(1 + r/n)^{(nt)}$ , where: $\newline$ $A$ = the future value of the investment/loan, including interest $\newline$ $P$ = the principal investment amount (the initial deposit or loan amount) $\newline$ $r$ = the annual interest rate (decimal) $\newline$ $n$ = the number of times that interest is compounded per year $\newline$ $t$ = the number of years the money is invested or borrowed for $\newline$ In this case, we have: $\newline$ $A = \$5000$ $\newline$ $P = \$500$ $\newline$ $r = 4.5\%$ or $0.045$ (as a decimal) $\newline$ $A$ $0$ (since the interest is compounded monthly) $\newline$ We need to find $t$ .
Convert Interest Rate: Convert the percentage interest rate to a decimal by dividing by $100$ . $\newline$ $r = \frac{4.5\%}{100} = 0.045$
Substitute and Solve: Substitute the known values into the compound interest formula and solve for $t$ . $\$5000 = \$500(1 + \frac{0.045}{12})^{(12t)}$
Isolate Compound Interest Factor: Divide both sides of the equation by $\$500$ to isolate the compound interest factor. $10 = (1 + 0.045/12)^{(12t)}$
Take Natural Logarithm: Take the natural logarithm of both sides to solve for the exponent $12t$ . $\newline$ $\ln(10) = \ln\left((1 + \frac{0.045}{12})^{12t}\right)$
Apply Power Rule: Use the power rule of logarithms, which states that $\ln(a^b) = b\cdot\ln(a)$ , to bring down the exponent on the right side of the equation. $\newline$ $\ln(10) = 12t \cdot \ln(1 + \frac{0.045}{12})$
Divide to Solve for t: Divide both sides by $(12 \cdot \ln(1 + \frac{0.045}{12}))$ to solve for t. $\newline$ $t = \frac{\ln(10)}{(12 \cdot \ln(1 + \frac{0.045}{12}))}$
Calculate t: Calculate the value of t using a calculator. $\newline$ $t \approx \frac{\ln(10)}{12 \times \ln(1 + \frac{0.045}{12})}$ $\newline$ $t \approx \frac{2.30258509299}{12 \times \ln(1.00375)}$ $\newline$ $t \approx \frac{2.30258509299}{12 \times 0.003731373}$ $\newline$ $t \approx \frac{2.30258509299}{0.044776476}$ $\newline$ $t \approx 51.4019185$
Interpret Result: Since $t$ represents the number of years, and we have a non-integer number of years, we need to interpret this result. It takes approximately $51.4$ years for the investment to grow from $\$500$ to $\$5000$ at an annual interest rate of $4.5\%$ , compounded monthly.