Money is deposited in an account for which the interest is compounded continuously. The initial investment in the account is $2000 and the annual interest rate is 14%. What is the time required for the balance to double? Round your answer to three decimal places.
To find the time required for the balance to double, we need to use the formula for continuous compound interest:
A = P * e^(rt)
Where:
A is the final amount in the account
P is the initial investment
e is Euler's number (approximately 2.71828)
r is the annual interest rate (as a decimal)
t is the time in years
In this case, the initial investment P is $2000 and the desired final amount A is $4000 (double the initial investment).
Let's plug in the given values into the formula and solve for t:
4000 = 2000 * e^(0.14t)
Dividing both sides by 2000:
2 = e^(0.14t)
Next, take the natural logarithm of both sides to cancel out the exponential term:
ln(2) = ln(e^(0.14t))
Using the property ln(x^a) = a * ln(x):
ln(2) = 0.14t * ln(e)
Since ln(e) is equal to 1, the equation simplifies to:
ln(2) = 0.14t
Now, divide both sides by 0.14 and solve for t:
t = ln(2) / 0.14
Using a calculator, evaluate this expression:
t ≈ 4.963
Therefore, the time required for the balance to double is approximately 4.963 years (rounded to three decimal places).