If $1000 is invested at a 3.5% interest rate, what is the least number of years until it is worth $1200?
I = PRT
200 = 1,000 * 0.035 * t
200 = 35t
5.714 = t
To find the least number of years until $1000 invested at a 3.5% interest rate is worth $1200, we can use the formula for compound interest:
\[A = P(1 + r/n)^(nt)\]
Where:
A is the final amount,
P is the principal (initial amount),
r is the annual interest rate (as a decimal),
n is the number of times interest is compounded per year,
and t is the number of years.
In this case, the principal (P) is $1000, the annual interest rate (r) is 3.5% or 0.035, and the amount we want to reach is $1200.
Since we are looking for the least number of years, we can start with small values of n (compounding frequency) to find the answer more quickly. Let's start with n = 1 (compounded annually).
The equation becomes:
\[1200 = 1000(1 + 0.035/1)^(1t)\]
Simplifying this equation:
\[1.2 = (1.035)^t\]
To solve for t (the number of years), we can take the logarithm of both sides:
\[log(1.2) = log((1.035)^t)\]
Using the logarithm property \(log(a^b) = b \cdot log(a)\):
\[log(1.2) = t \cdot log(1.035)\]
Rearranging the equation:
\[t = \frac{log(1.2)}{log(1.035)}\]
Using a calculator, we can calculate the value of t by dividing the natural logarithm of 1.2 (ln(1.2)) by the natural logarithm of 1.035 (ln(1.035)). The result is approximately 7.79.
Therefore, the least number of years until $1000 invested at a 3.5% interest rate is worth $1200 is approximately 7.79 years (rounded to the nearest whole number).