How long does it take for an amount of money P to double itself if its invested at 8℅ interest compounded 4 times a year? Round your answer to 2 decimal places.
(1+.08/4)^(4t) = 2
t = 8.75 years
To determine how long it takes for an amount of money P to double itself with compound interest, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the final amount (double the initial amount in this case)
P = the principal amount (initial amount)
r = annual interest rate (in decimal form)
n = number of times interest is compounded per year
t = number of years
In this case, P is the initial amount that will double, r is 8% (which is equivalent to 0.08 in decimal form), n is 4 (interest compounded 4 times a year), and A is 2P (double the initial amount).
2P = P(1 + 0.08/4)^(4t)
Now we can solve for t. Divide both sides of the equation by P:
2 = (1 + 0.08/4)^(4t)
Take the natural logarithm of both sides to get rid of the exponent:
ln(2) = ln((1 + 0.08/4)^(4t))
Using the property of logarithms, we can bring down the exponent:
ln(2) = 4t * ln(1 + 0.08/4)
Simplify the right side:
ln(2) = 4t * ln(1.02)
Now divide both sides by 4ln(1.02):
t = ln(2) / (4 * ln(1.02))
Using a calculator or a programming language, we can evaluate this expression:
t ≈ 8.66 years
Therefore, it will take approximately 8.66 years for the money to double itself at an interest rate of 8% compounded 4 times a year.