Final answer:
By using the compound interest formula, it is determined that Araba's $3000 deposit at an annual interest rate of 1% will take approximately 69.66 years to double in value.
Step-by-step explanation:
The student's question concerns the calculation of the time it will take for an investment to double when it is subjected to compound interest. Specifically, Araba wants to know how long it will take for her $3000 checking account deposit to double at an annual interest rate of 1%. To solve this, we can use the compound interest formula:
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 (decimal).
- n is the number of times that interest is compounded per year.
- t is the time in years.
We want the final amount A to be double the initial deposit, so A = 2P. Since the interest is credited annually, n is 1. For Araba's case, where P = $3000 and r = 0.01 (1%), we wish to find t such that:
2 * 3000 = 3000(1 + 0.01/1)^(1*t)
This simplifies to:
2 = (1.01)^t
To solve for t, we can take the natural logarithm of both sides:
ln(2) = t * ln(1.01)
Therefore, t = ln(2) / ln(1.01)
Calculating this value gives us:
t ≈ 69.66 years
So, Araba's deposit will take approximately 69.66 years to double in her checking account.