question below....

Question

A bank pays r=5% interest compounded monthly. if you deposit P0 dollars at a time t=0, then the value of your account after N months is P0 (1+(0.05/12))^N. find to the nearest integer N the number of months after which the account value doubles.... I believe the answer is 0 because it seems what I'm getting out of the question you have no money in or going into your account. am I right? if not can you please explain?

Answer 1

A = P_o(1 + 0.05/12)^N

 

You want to determine when A = 2P_o

 

2P = P(1 + 0.05/12)^N                     divide both sides by P

2 = (1 + 0.0041666)^N

2 = 1.0041666)^N                            take ln of both sides

ln 2 ≈ N ln (1.0041666)

ln 2 ≈ (0.00415794376) N

ln 2 / (0.00415794376) ≈ N

0.693147181 / 0.00415794376 ≈ N

166.7 ≈ N

 

The principal doubles in 166.7 months.

Answer 2

Your formula appears to be missing an element.

 

The compound interest formula is:

A = P₀(1+r/n)^nt

A = future or final amount

P₀= principal (initial investment) = .05

r = interest rate as a decimal

n = # times per year interest os compounded=12

t = time in years

 

For this problem we want to know how long until A = 2P0

2P₀=P₀(1+.05/12)^12t

2=(1+.05/12)^12t

log2 = log[(1+.05/12)^12t)]

log2=(12t)log(1+.05/12)

t= log2/[12log(1+.05/12)]= 13.89 years

 

(13.89 years)(12 mos/yr)= 166.7 months

 

 

Answer 3

2P = P(1 + 0.05/12)^N, where P is the amount originally invested

 

P cancels out, then take the natural log of both sides to get                           

N = ln2/(0.00415794376) = 166.7 months (or about 13.9 years)