Question

$4000 is invested in a bank account at an interest rate of 8% per year, compounded continuously. Meanwhile, $17000 is invested in a bank account at an interest rate of 3% compounded annually. To the nearest year, When will the two accounts have the same balance? After how many years?

Answer 1

well continuous compounding is

\[A = P \times e^{rt}\]

A = future value, P = principal r = interest rate as a decimal and t = time

plug the values in

Answer 2

im wasnt sure which one to use

Answer 3

the annual compound interest formula is

\[A = P \times (1 + r)^t\]

the letter have the same definitions...

just plug the numbers in

Answer 4

but one of them says comopunded annually

Answer 5

so then you need to equate both formulae and then solve for t. to find the time when the balances are the same..

Answer 6

so for the first one i got 4000e^(.08*t)

Answer 7

is that part right?

Answer 8

that's correct

Answer 9

so the second equation is 17000(1.03)^t right?

Answer 10

that's correct... so the equation becomes

\[4000e^{0.8t} = 17000(1.03)^t\]

divide by 4000

\[e^{0.8t} = 4.25\times (1.03)^t\]

now you need to take the log of both sides...

so use base e logs or ln..

hope it helps

Answer 11

oh ok i get it, its cuz when i divided by 4000 i put it under (1.03)^t as well

Answer 12

you well need to know about

1. log laws for multiplication

2. log laws for powers

3. log laws for the log of the base...

Answer 13

so now you should have 0.08t = ln4.25(1.03)^t right?

Answer 14

that's correct... so the law for multiplication says add the logs

\[0.08t = \log(4.25) + \log(1.03)^t\]

now the law for powers

\[0.08t = \ln(4.25) + t \ln(1.03)\]

just collect the like terms and sole for t

Answer 15

so when your adding you put t(ln(4.25 * 1.03)) right?

Answer 16

collect like terms

0.08t - t(ln(1.03)) = ln(4.25)

factor the left

t(0.08 - ln(1.03) ) = ln(4.25)

now solve for t

Answer 17

oh ok i get it thnx....i see what i did wrong