Suppose you have $\$ 7500$ deposited at $2.35 \%$ compounded daily. About long will it take your balance to increase to $\$ 9000$ ?

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Q. Suppose you have

\$ 7500

deposited at

2.35 \%

compounded daily. About long will it take your balance to increase to

\$ 9000

Use Compound Interest Formula: First, we need to use the formula for compound interest which 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. $\newline$ We are given: $\newline$ $A = \$9000$ $\newline$ $P = \$7500$ $\newline$ r = $2.35\%$ or $0.0235$ in decimal $\newline$ $n = 365$ (since the interest is compounded daily) $\newline$ We need to find $t$ .
Convert Interest Rate to Decimal: Convert the percentage interest rate to a decimal by dividing by $100$ : $\newline$ $r = 2.35\% = \frac{2.35}{100} = 0.0235$
Rearrange Formula to Solve for t: Now we will rearrange the compound interest formula to solve for t: $\newline$ $9000 = 7500(1 + \frac{0.0235}{365})^{365t}$
Isolate Compound Interest Factor: Divide both sides by $\$7500$ to isolate the compound interest factor: $\newline$ $\$9000 / \$7500 = (1 + 0.0235/365)^{365t}$ $\newline$ $1.2 = (1 + 0.0235/365)^{365t}$
Take Natural Logarithm of Both Sides: Take the natural logarithm (ln) of both sides to solve for the exponent: $\newline$ $\ln(1.2) = \ln((1 + 0.0235/365)^{365t})$ $\newline$ $\ln(1.2) = 365t \cdot \ln(1 + 0.0235/365)$
Calculate $\ln(1.2)$ and $\ln(1 + 0.0235/365)$ : Now we need to calculate $\ln(1.2)$ and $\ln(1 + 0.0235/365)$ using a calculator: $\newline$ $\ln(1.2) \approx 0.1823$ $\newline$ $\ln(1 + 0.0235/365) \approx \ln(1 + 0.00006438356) \approx 0.000064373$
Solve for t: Now we can solve for t: $\newline$ $0.1823 = 365t \times 0.000064373$ $\newline$ $t = \frac{0.1823}{365 \times 0.000064373}$ $\newline$ $t \approx 7.7$ years
Use More Precise $\ln(1 + 0.0235/365)$ : We will use a more precise value for $\ln(1 + 0.0235/365)$ using a calculator: $\ln(1 + 0.0235/365) \approx \ln(1.00006438356) \approx 0.000064348$
Solve for t with Corrected Value: Now we can solve for t with the corrected value: $\newline$ $0.1823 = 365t \times 0.000064348$ $\newline$ $t = \frac{0.1823}{(365 \times 0.000064348)}$ $\newline$ $t \approx 7.7$ years