Calculus help please!!!! Compound Interest?

4000 dollars is invested in a bank account at an interest rate of 8 per cent per year, compounded continuously. Meanwhile, 15000 dollars is invested in a bank account at an interest rate of 2 percent compounded annually.

To the nearest year, When will the two accounts have the same balance?

The two accounts will have the same balance after ____ years.

Calculus help please!!!! Compound Interest?cheap loans

Let t be the number of years.

Equal the balance after t years,

4000e^.08t = 15000(1.02)^t, where the LHS is the one compounded continuously, while the RHS is the one compounded annually.

Collect variables in one side,

(e^.08/1.02)^t = 15/4

Take ln, and solve for t,

t = ln(15/4)/ln(e^.08/1.02) = 21.96 ~ 22 years

So, the two accounts will have the same balance after 22 years.

Calculus help please!!!! Compound Interest?

loan

17 years|||I don%26#039;t know what continuously means... but the formula is:

E=S*P^t/c

E= End amount

S= Start amount

P= Percent %26lt;or%26gt; Interest Rate

t = total time

c = time for one cycle (ex. compounded monthy would be 1/12)

^= To the power of

*= multiplied by

/= divided by

Hope this helps|||Compounded contiuously means that you have to use the equation A = Pe^rt. You will substitute P as the dollars invest, r as the percent interest, and t is what you are trying to find. Set the two equations equal to each other, which means:

4000e^(.08t) = 15000e^(.02t) and you can figure this out by taking the natural log on both sides. The nearest year for this would be 22 years. You can substitute in 22 for t in both different equations and the amounts will be very close (this is because it is asking for the nearest year so 22 is a rounded number)