Question
Stephanie has a customer that is making a one-time $10, 000 deposit into a new account. The
account has a 4.5% annual interest rate compounded monthly. The customer makes no other
deposits or withdrawals.
In about how many years will the customer have $12, 500 in the account?
Hint: The student understands how to create an exponential equation and use logarithms to
solve applicable problems. The student understands that P = 10, 000, A = 12, 500, r = 0.045,
n = 12, and t as the unknown. [ A = P (1 +
π
π
)
ππ‘ ]
A) 1 year
B) 5 years
C) 12 years
D) 15 years
account has a 4.5% annual interest rate compounded monthly. The customer makes no other
deposits or withdrawals.
In about how many years will the customer have $12, 500 in the account?
Hint: The student understands how to create an exponential equation and use logarithms to
solve applicable problems. The student understands that P = 10, 000, A = 12, 500, r = 0.045,
n = 12, and t as the unknown. [ A = P (1 +
π
π
)
ππ‘ ]
A) 1 year
B) 5 years
C) 12 years
D) 15 years
Answers
GPT-4o mini
To calculate how long it will take for the account to grow from $10,000 to $12,500 with a 4.5% annual interest rate compounded monthly, we can use the formula:
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \(A\) is the amount of money accumulated after n years, including interest.
- \(P\) is the principal amount (the initial deposit).
- \(r\) is the annual interest rate (decimal).
- \(n\) is the number of times that interest is compounded per year.
- \(t\) is the number of years the money is invested or borrowed for.
Given:
- \(A = 12,500\)
- \(P = 10,000\)
- \(r = 0.045\)
- \(n = 12\)
We need to solve for \(t\):
\[
12500 = 10000 \left(1 + \frac{0.045}{12}\right)^{12t}
\]
First, simplify the fraction inside the parentheses:
\[
1 + \frac{0.045}{12} = 1 + 0.00375 = 1.00375
\]
Now rewrite the equation:
\[
12500 = 10000 (1.00375)^{12t}
\]
Next, divide both sides by 10,000:
\[
1.25 = (1.00375)^{12t}
\]
Next, take the logarithm of both sides. Using the natural logarithm (ln) for simplicity:
\[
\ln(1.25) = \ln((1.00375)^{12t})
\]
Using the property of logarithms that allows us to bring the exponent in front:
\[
\ln(1.25) = 12t \cdot \ln(1.00375)
\]
Now, solve for \(t\):
\[
t = \frac{\ln(1.25)}{12 \cdot \ln(1.00375)}
\]
Now, calculate the values:
First, calculate \(\ln(1.25)\):
\[
\ln(1.25) \approx 0.22314
\]
Next, calculate \(\ln(1.00375)\):
\[
\ln(1.00375) \approx 0.003738
\]
Now substitute back into the equation for \(t\):
\[
t \approx \frac{0.22314}{12 \cdot 0.003738}
\]
\[
t \approx \frac{0.22314}{0.044856} \approx 4.973 \text{ years}
\]
This rounds to approximately 5 years. Therefore, the answer is:
**B) 5 years**.
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \(A\) is the amount of money accumulated after n years, including interest.
- \(P\) is the principal amount (the initial deposit).
- \(r\) is the annual interest rate (decimal).
- \(n\) is the number of times that interest is compounded per year.
- \(t\) is the number of years the money is invested or borrowed for.
Given:
- \(A = 12,500\)
- \(P = 10,000\)
- \(r = 0.045\)
- \(n = 12\)
We need to solve for \(t\):
\[
12500 = 10000 \left(1 + \frac{0.045}{12}\right)^{12t}
\]
First, simplify the fraction inside the parentheses:
\[
1 + \frac{0.045}{12} = 1 + 0.00375 = 1.00375
\]
Now rewrite the equation:
\[
12500 = 10000 (1.00375)^{12t}
\]
Next, divide both sides by 10,000:
\[
1.25 = (1.00375)^{12t}
\]
Next, take the logarithm of both sides. Using the natural logarithm (ln) for simplicity:
\[
\ln(1.25) = \ln((1.00375)^{12t})
\]
Using the property of logarithms that allows us to bring the exponent in front:
\[
\ln(1.25) = 12t \cdot \ln(1.00375)
\]
Now, solve for \(t\):
\[
t = \frac{\ln(1.25)}{12 \cdot \ln(1.00375)}
\]
Now, calculate the values:
First, calculate \(\ln(1.25)\):
\[
\ln(1.25) \approx 0.22314
\]
Next, calculate \(\ln(1.00375)\):
\[
\ln(1.00375) \approx 0.003738
\]
Now substitute back into the equation for \(t\):
\[
t \approx \frac{0.22314}{12 \cdot 0.003738}
\]
\[
t \approx \frac{0.22314}{0.044856} \approx 4.973 \text{ years}
\]
This rounds to approximately 5 years. Therefore, the answer is:
**B) 5 years**.