Suppose that $3000 is placed in a savings account at an annual rate of 6%, compounded monthly. Assuming that no wi for the account to grow to $4818? Do not round any intermediate computations, and round your answer to the nearest hundredth.
Suppose that $3000 is placed in a savings account at an annual rate of 6%, compounded monthly. Assuming that no wi for the account to grow to $4818? Do not round any intermediate computations, and round your answer to the nearest hundredth.
Chapter1: Financial Statements And Business Decisions
Section: Chapter Questions
Problem 1Q
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![# Compound Interest Calculation Example
**Problem Statement:**
Suppose that $3000 is placed in a savings account at an annual rate of 6%, compounded monthly. Assuming that no withdrawals are made, how long will it take for the account to grow to $4818?
Do not round any intermediate computations, and round your answer to the nearest hundredth.
**Solution Steps:**
1. **Identify the given values:**
- Principal amount (P): $3000
- Annual interest rate (r): 6% (or 0.06 when expressed as a decimal)
- Final amount (A): $4818
- Number of times interest is compounded per year (n): 12 (since it is compounded monthly)
2. **Formula for compound interest:**
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \(A\) is the amount of money accumulated after n years, including interest.
- \(P\) is the principal amount (the initial amount of money).
- \(r\) is the annual interest rate (decimal).
- \(n\) is the number of times that interest is compounded per unit time.
- \(t\) is the time the money is invested for, in years.
3. **Plugging in the given values into the formula:**
\[
4818 = 3000 \left(1 + \frac{0.06}{12}\right)^{12t}
\]
4. **Solving for t (time in years):**
\[
\frac{4818}{3000} = \left(1 + \frac{0.06}{12}\right)^{12t}
\]
\[
1.606 = \left(1 + 0.005\right)^{12t}
\]
\[
1.606 = 1.005^{12t}
\]
5. **Taking the natural logarithm on both sides to solve for \(t\):**
\[
\ln(1.606) = \ln(1.005^{12t})
\]
\[
\ln(1.606) = 12t \cdot \ln(1.005)
\]
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Transcribed Image Text:# Compound Interest Calculation Example
**Problem Statement:**
Suppose that $3000 is placed in a savings account at an annual rate of 6%, compounded monthly. Assuming that no withdrawals are made, how long will it take for the account to grow to $4818?
Do not round any intermediate computations, and round your answer to the nearest hundredth.
**Solution Steps:**
1. **Identify the given values:**
- Principal amount (P): $3000
- Annual interest rate (r): 6% (or 0.06 when expressed as a decimal)
- Final amount (A): $4818
- Number of times interest is compounded per year (n): 12 (since it is compounded monthly)
2. **Formula for compound interest:**
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \(A\) is the amount of money accumulated after n years, including interest.
- \(P\) is the principal amount (the initial amount of money).
- \(r\) is the annual interest rate (decimal).
- \(n\) is the number of times that interest is compounded per unit time.
- \(t\) is the time the money is invested for, in years.
3. **Plugging in the given values into the formula:**
\[
4818 = 3000 \left(1 + \frac{0.06}{12}\right)^{12t}
\]
4. **Solving for t (time in years):**
\[
\frac{4818}{3000} = \left(1 + \frac{0.06}{12}\right)^{12t}
\]
\[
1.606 = \left(1 + 0.005\right)^{12t}
\]
\[
1.606 = 1.005^{12t}
\]
5. **Taking the natural logarithm on both sides to solve for \(t\):**
\[
\ln(1.606) = \ln(1.005^{12t})
\]
\[
\ln(1.606) = 12t \cdot \ln(1.005)
\]
\
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