Question

I have been using every math formula I know to figure this out what am I doing wrong? I would really appreciate some help. thanks :)

Kate currently has an account balance of $7,194.66. She opened the account 21 years ago with a deposit of $2,978.41. If the interest compounds daily, what is the interest rate on the account?

Answer 1

Use
\[\large A=P(1+R)^t\]

Answer 2

In this case, since the compounding is daily, you have to convert 21 years to days for 't'

Answer 3

alright thanks im gonna try to see if this works thanks!

Answer 4

so would I divide 21 by 7?

Answer 5

Number of days in 21 years = (21years) × (365.25days/year).

Answer 6

Also remember that since this is being solved in terms of days, you need to multiply the R that you get by 365.25 to get it in terms of years (APR).

Answer 7

okey dokey thanks !!! :)

Answer 8

im wondering if "compound daily" is the same as "continuous"

Answer 9

hmmm.... then I should use the formula A=Pe^rt

Answer 10

Pert would be my idea
\[A=Pe^{rt}\]
\[ln(\frac{A}{P})=rt\]
\[\frac{1}{t}ln(\frac{A}{P})=r\]

Answer 11

ok im trying that thanks.

Answer 12

"compound daily" is not the same as "continuous"...though it would prob be a good approximation

Answer 13

ok so it isnt the same? im now confused.

Answer 14

daily=once a day interest is compounded
continuous=every instant interest is compounded

Answer 15

ahhh ok i get it now thank you :)

Answer 16

Daily gives me: .0X2
Continuous gives me: .0X19976....

so yeah, they are pretty close

Answer 17

Continuous compounding is a good approximation, but it would be good algebra practice to actually do it with daily.

Answer 18

I'd start with
\[\large A=P(1+R)^T\]
\[\large \rightarrow log(A/P)=T \cdot log(1+R)\]
\[\large let \space X=log(A/P)/T \rightarrow R=10^X-1\]
That R is the daily rate using T=7670.25 days, so multiply by 365.25 to get the APR.

Answer 19

alright then thanks everyone!!! :)