Please help with this calc question?

4000 dollars is invested in a bank account at an interest rate of 8% per year, compounded continuously. Meanwhile, 11000 dollars is invested in a bank account at an interest rate of 5% compounded annually.

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

Please help with this calc question?loan company

this is the equation for a continuously compounded investment:

A = Pe^(rt)

where

A = total amount (current worth)

P = initial deposit (or principal)

r = annual interest rate (expressed as a fraction)

t = number of years invested

so, using that equation and plugging in the numbers:

A = 4000 e^(0.08t)

this is the equation for an annually compounded investment:

A = P (1 + r)^t

so, using that equation and plugging in the numbers:

A = 11000 (1 + 0.05)^t

now, we set those two equations equal to each other:

4000e^(0.08t) = 11000(1.05)^t

divide both sides by 4000:

e^0.08t = 2.75 (1.05)^t

take the natural log of both sides:

ln (e^0.08t) = ln (2.75 (1.05)^t)

thus:

0.08t = ln (2.75 (1.05)^t)

0.08t = ln (2.75) + ln (1.05^t)

0.08t = ln (2.75) + t (ln (1.05))

collect the terms with t to one side:

0.08t - t (ln (1.05)) = ln (2.75)

t (0.08 - ln (1.05)) = ln (2.75)

divide both sides by (0.08 - ln (1.05)) to solve for t:

t = (ln (2.75)) / (0.08 - ln (1.05))

plug that into your calculator to get an actual number:

t = 32.413 years

i hope that helped!