# Thread: Help with simple financial problem

1. ## Help with simple financial problem

Suppose on Jan 1, 1997 Dave invested $2,000 into a bank account at 5% interest compounded continuously. Let y(t) be the value of Dave's investment after t years. Also on Jan 1, 1997 John decides to invest. He put$2,500 into an account at
3% interest compounded monthly. Let g(t) be the value of John's investment after t years.

To the nearest tenth, at what time t is the value of both accounts the same?

So I know you set ;

2000e^(.03)t = 2,500 (1 + .03/12) ^ (12)t

How do you go about solving?

2. Take the log of both sides, and then use the fact that
$ln (a^b) = b (ln (a))$

3. Originally Posted by nikkkk
Suppose on Jan 1, 1997 Dave invested $2,000 into a bank account at 5% interest compounded continuously. Let y(t) be the value of Dave's investment after t years. Also on Jan 1, 1997 John decides to invest. He put$2,500 into an account at
3% interest compounded monthly. Let g(t) be the value of John's investment after t years.

To the nearest tenth, at what time t is the value of both accounts the same?

So I know you set ;

2000e^(.03)t = 2,500 (1 + .03/12) ^ (12)t

How do you go about solving?
If the values for t had the same base this would be very easy to solve.

I would put these functions into excel and find where they are equal.