Continuous Compound Interest Formula

Question 3

An amount of $3,000.00 is deposited in a bank paying an annual interest rate of 3 %, compounded continuously.

(a) Find the balance after 4 years.
(b) How long would it take for the money to double?

Solution

Use the continuous compound interest formula,
A = Pe rt, with P = 3000, r = 3/100 = 0.03, t = 4.

(a) Therefore,

Solution to Question 3

So, the balance after 4 years is approximately $3,382.49.


(b) Since the original investment is $3,000.00, doubling means that the current balance is $6,000.00. To find out how long it takes for this to happen ( i.e. to find t ), plug in P = 3000, A = 6000, and r = 0.03 in the continuous compound interest formula, and solve for t. Doing this, one gets,

  3000 e 0.03 t = 6000
e 0.03 t = 6000/3000 = 2		  

So we have to solve the exponential equation, e 0.03 t = 2, by converting it into log notation. This will give us,

  0.03t = loge2
0.03t = ln2
t = ln2 / 0.03
t = 23.10490601866		  

Therefore, it would take approximately 23.1 years for the money to double.

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