You invest 3,500 in a savings account with a 4.5% annual interest rate, and your account balance grows to 4,130. How many years did it take for your investment to reach this balance?
Question
You invest 3,500 in a savings account with a 4.5% annual interest rate, and your account balance grows to 4,130. How many years did it take for your investment to reach this balance?
Solution
To solve this problem, we can use the formula for compound interest, which is A = P(1 + r/n)^(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 (in decimal form, so 4.5% becomes 0.045).
- n is the number of times that interest is compounded per year.
- t is the time the money is invested for in years.
In this case, we are trying to solve for t (time), so we need to rearrange the formula to solve for t. The rearranged formula is t = log(A/P) / n[log(1 + r/n)].
Given that the interest is likely compounded annually (n=1), the formula simplifies to t = log(A/P) / log(1 + r).
Substituting the given values into the formula:
- A = 4,130
- P = 3,500
- r = 0.045
We get t = log(4130/3500) / log(1 + 0.045).
Calculating the values, we get t = 0.0641 / 0.0431, which gives t ≈ 1.49 years.
So, it took approximately 1.49 years for the investment to grow from 3,500 to 4,130 with an annual interest rate of 4.5%.
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