Question 6356
A=$6000 P= $1000 interest is 12% compounded quarterly
Find # of interest periods
`<font face = "courier new" color = "blue"><b>
The formula is 
`
A = P(1+r/n)<sup>nt</sup>
`
Where
`
A = ending amount = 6000
P = beginning amount = 1000
r = annual rate expressed as a decimal = .12
n = number of interest periods per year = 4
` ` (since it's quarterly)
t = number of years  (which we find first)
`
A = P(1+r/n)<sup>nt</sup>
`
Taking logs of both sides
`
logA = log[P(1+r/n)<sup>nt</sup>]
`
logA = logP + log(1+r/n)<sup>nt</sup>
`
logA = logP + nt[log(1+r/n)]
`
logP + nt[log(1+r/n)] = logA 
`
nt[log(1+r/n)] = logA - logP
`
nt[log(1+r/n)] `` logA - logP
-------------- = -------------
`n[log(1+r/n)] ` n[log(1+r/n)] 
`
nt[log(1+r/n)] `` logA - logP
-------------- = -------------
`n[log(1+r/n)] ` n[log(1+r/n)]
`
` ` ` ` ` ` ` ``` logA - logP
` ` ` ` ` ` `t = -------------
` ` ` ` ` ` ` `` n[log(1+r/n)]
`
` ` ` ` ` ` ` ``` log6000 - log1000
` ` ` ` ` ` `t = --------------------
` ` ` ` ` ` ` ` ` `4[log(1+.12/4)]
`
` ` ` ` ` ` ` ``` 3.77815125 - 3
` ` ` ` ` ` `t = --------------------
` ` ` ` ` ` ` ` ` `4[log(1.03)]
`
` ` ` ` ` ` ` `` ` .7781512504 
` ` ` ` ` ` `t = ---------------
` ` ` ` ` ` ` ` ` `.0513488988
`
` ` ` ` ` ` ` `` ` .77815125 
` ` ` ` ` ` `t = ---------------
` ` ` ` ` ` ` ` ` `.0513488988
`
` ` ` ` ` ` `t =  15.15419548 years
`
Multiply by 4 to get the number of interest periods
`
Answer = 60.61678192 which means that 60 interest periods will
yield slightly less that $6000, or $5891.60 and 61 interest period
will yield slightly more than $6000, or $6068.40.
`
Edwin <font face = "wingdings" color = "red" size = 7>J</font></b>