Continuous Compound Interest Formula

Question 2

An amount of $2,000.00 is deposited in a bank paying an annual interest rate of 2.85 %, compounded continuously.

(a) Find the balance after 3 years.
(b) How long would it take for the money to double?

Solution

Use the continuous compound interest formula,
A = Pe rt, with P = 2000, r = 2.85/100 = 0.0285, t = 3.

(a) Therefore,

Solution to Question 2

So, the balance after 3 years is approximately $2,178.52.


(b) Since the original investment is $2,000.00, doubling means that the current balance is $4,000.00. To find out how long it takes for this to happen ( i.e. to find t ), plug in P = 2000, A = 4000, and r = 0.0285 in the continuous compound interest formula, and solve for t. Doing this, one gets,

  2000 e 0.0285 t = 4000
e 0.0285 t = 4000/2000 = 2		  

So we have to solve the exponential equation, e 0.0285 t = 2, by converting it into log notation. This will give us,

  0.0285t = loge2
0.0285t = ln2
t = ln2 / 0.0285
t = 24.32095370386		  

Therefore, it would take approximately 24.32 years for the money to double.

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