Question

You deposit $2000 in Account A, which pays 2.25% annual interest compounded monthly. You deposit another $2000 in Account B, which pays 3% annual interest compounded monthly. When is the sum of the balance in both accounts at least $5000?

Answer 1

Nice username.

Answer 2

hmmm I assume you've used the compound interest formula already?

Answer 3

yup. Im on the chapter using logarithms and exponential functions but it doesnt seem to work
I know the answer is supposed to be 8.5 if that helps. I just dont know how to do it

Answer 4

hmm

Answer 5

So basically can you tell me the formulas in compounded interest form
Note: make x in years, so in Account B you'll need to put x/12

Answer 6

my formula was:
2000(1+0.0225/12)^12x + 2000(1+0.03/12)^12x is greater than or equal to 5000

Answer 7

the cycle is yearly, so the cycle or "n = 1" in this case
so \(\bf A=P\left(1+\frac{r}{1}\right)^{1t}\implies A=P(1+r)^t\)

so.. the two balances need to be greater than or equal to 5000
so pretty much \(\bf 5000 \ge A + A\)

Answer 8

lemme post what I have so far

Answer 9

ok thanks
i thought since it said compounded monthly it meant n=12 ... thats what we have been doing in class

Answer 10

ohmmm shoot.. lemme reread that

Answer 11

rats, yes, I read yearly smokes... yes ...so n = 12 ok

Answer 12

\[2000(1+0.0225/12)^{12x} + 2000(1+0.03/12)^{12x}\] Ah yes I think I;ve got it
so take the log of both sides, ignore the greater than make it just equals for now\[2000(1+0.0225/12)^{12x} + 2000(1+0.03/12)^{12x}=5000\]

Answer 13

Take the log or ln or whatever you want of both sides which makes it\[\log_{10} (2000(1+0.0225/12)^{12x} + 2000(1+0.03/12)^{12x})\]

Answer 14

Sorry
\[\log10(2000(1+0.0225/12)^{12x}+2000(1+0.03/12)^{12x})=\log_{10}5000 \]

Answer 15

Ahhh wait no can't do that

Answer 16

lemme show you what my teacher did in class

Answer 17

Wait let's restart\[2000(1+0.0225/12)^{12x} + 2000(1+0.03/12)^{12x}=5000\]
Divide both sides by 2000 you get\[(12.0225)^{12x}+(1.03)^{12x}=2.5\]

Answer 18

OHH Ok I see what your teacher did
ALthough back up two steps that way takes you backward

Answer 19

alrighty then :)

Answer 20

Wait no
Didn't see that x
Hmm

Answer 21

Ok yeah I honestly have no clue
Either route I take mine or your teachers I'm at a dead end

Answer 22

Sorry :/

Answer 23

lol thats ok, even my teacher couldnt figure it out
thanks for the effort though!!

Answer 24

The problem reduces to the following, where t is the time in years:
\[\large 1.001875^{12t}+1.0025^{12t}=2.5\]
A value of t = 8.75 years is a close solution.

Answer 25

yeah i got that reduced equation but how did you come to that answer?
did u just do guess and check?

Answer 26

I calculated the value of both terms on the left starting with 12t = 100. It soon emerges that12t = 105 gives a good approximation.

Answer 27

alright thanks!

Answer 28

You're welcome :)

Answer 29

checking this or that way.... I don't see how to get the logs though to tweeze out the "t"

Answer 30

yeah any way i try it the t is still in the exponent somewhere

Answer 31

Yeah pretty sure the only ways to get an answer here is guessing and checking or graphing