Question

Suppose one bank account pays 6 % annual interest compounded once per year, and a second bank account pays 6 % annual interest compounded continuously. If both bank accounts start with the same initial amount, how long will it take for the second bank account to contain twice the amount of the first bank account?

Round your answer to the nearest year.

Answer 1

Let A1 be the amount after t years in the account where interest is compounded once per year.
Let A2 be the amount after t years in the account where interest is compounded continuously.
\[\large \frac{Pe^{0.06t}}{P(1+0.06)^{t}}=\frac{A _{2}}{A _{1}}=2\ ...........(1)\]
Cross multiplying in equation (1) we get:
\[\large Pe^{0.06t}=2P(1.06)^{t}\ .........(2)\]
After dividing both sides of (2) by P we get:
\[\large e^{0.06t}=2\times1.06^{t}\ ............(3)\]
Taking natural logs of both sides of (3) results in:
\[\large 0.06t=\ln 2+t \ln 1.06\ .............(4)\]
Now you just need to solve equation (4) to find the value of t.