How long do you need to invest your money in an account earning an annual interest rate of 3.069% compounded monthly so that your investment grows from $1426.43 to $11980.00 over that period of time?
1426.43(1 + .03069/12)^n = 1198.00
Solve for n, the number of months.
Thank you for your answer
How can I solve it for N?
What is the way?
To determine how long you need to invest your money, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the future value of the investment
P = the initial principal (starting amount)
r = the annual interest rate (expressed as a decimal)
n = the number of times interest is compounded per year
t = the time period (in years)
In this case, we are given the following information:
A = $11,980.00 (the desired future value of the investment)
P = $1,426.43 (the initial principal)
r = 3.069% per year (or 0.03069 as a decimal)
n = 12 (compounded monthly)
We want to find t, the time period in years. Rearranging the formula, we have:
t = (1/n) * log(A/P) / log(1 + r/n)
Let's calculate it step by step.
Step 1: Convert the annual interest rate to a monthly interest rate:
Monthly interest rate = r/n = 0.03069 / 12 = 0.00257
Step 2: Substitute the given values into the formula:
t = (1/12) * log(11,980.00 / 1,426.43) / log(1 + 0.00257)
Step 3: Simplify the equation:
t = 0.0833 * log(8.38465) / log(1.00257)
Step 4: Calculate the logarithms using a calculator:
t ≈ 0.0833 * 0.9234 / 0.0009
Step 5: Calculate the final result:
t ≈ 0.0763 years
Therefore, you would need to invest your money for approximately 0.0763 years, which is equivalent to approximately 9.2 days, for your investment to grow from $1,426.43 to $11,980.00.