Question

Phillip deposited $7,775 into a savings account 14 years ago. The account has an interest rate of 4.5% and the balance is currently $14,398.87. How often does the interest compound?

Answer 1

\[\Large A = P(1 + \frac{ r }{ n })^{nt}\]A = Maturity Amount = $14,398.87
P = Principal = $7,775
r = rate of interest in decimal = 4.5/100 = 0.045
t = time duration of deposit = 14 years
n = compounding period

Plug in all the values and solve for n.

Answer 2

okay so A=14,398.87(1+0.045/n)^n14

Answer 3

A=14,398.87
P = $7,775

Answer 4

OHH crap
14,398.87=7,775(1+0.045/n)^n14

Answer 5

Divide both sides by 7775

Answer 6

16.85=(1+0.045/n)^n14

Answer 7

Do you have a financial calculator or a TVM table in the book that you can look up?

Answer 8

no i do not /:

Answer 9

I am trying to search for ways to solve it without those two.

Answer 10

You can take log on both sides and then divide both sides by 14.

Answer 11

1.2=(1+0.045/n)^n ?

Answer 12

When you take log on both sides it becomes:
1.23 = 14n * log(1+0.045/n)
Divide both sides by 14
0.0876 = n * log(1 + 0.045/n)
This is a difficult equation to solve without a table to lookup or the proper calculator.
But trial and error might work. Try n = 1, 2, 3, 4, etc. and see which one satisfies the equation.

Answer 13

Okay well thank you for your help!

Answer 14

Wait, how did you get 16.85 = (1+0.045/n)^n14 ?

Answer 15

14,398.87=7,775(1+0.045/n)^(14n)
Divide both sides by 7775
1.8519 = (1+0.045/n)^(14n)
Take log on both sides
log(1.8519) = log{ (1+0.045/n)^(14n) } = 14n * log(1+0.045/n)
0.2676 = 14n * log(1+0.045/n)
Divide both sides by 14
0.0191 = n * log(1+0.045/n)

Trial and error method with n = 1, 2, 3, etc.
Try n = 1 and evaluate the right hand side:
1 * log(1 + 0.045) = log(1.045) = 0.0191
This agrees with our left hand side and we got the answer in the very first attempt!

So n = 1 which means the interest is compounded once every year.

Answer 16

ohhh okay! thank you! :D

Answer 17

you are welcome.

Answer 18

Annually!