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Question 6356: A=$6000 P= $1000 interest is 12% compounded quarterly Find # of interest periods
Answer by AnlytcPhil(1810)   (Show Source):
You can put this solution on YOUR website! A=$6000 P= $1000 interest is 12% compounded quarterly
Find # of interest periods
`
The formula is
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A = P(1+r/n)nt
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Where
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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)
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A = P(1+r/n)nt
`
Taking logs of both sides
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logA = log[P(1+r/n)nt]
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logA = logP + log(1+r/n)nt
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logA = logP + nt[log(1+r/n)]
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logP + nt[log(1+r/n)] = logA
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nt[log(1+r/n)] = logA - logP
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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)]
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` ` ` ` ` ` ` ``` logA - logP
` ` ` ` ` ` `t = -------------
` ` ` ` ` ` ` `` n[log(1+r/n)]
`
` ` ` ` ` ` ` ``` log6000 - log1000
` ` ` ` ` ` `t = --------------------
` ` ` ` ` ` ` ` ` `4[log(1+.12/4)]
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` ` ` ` ` ` ` ``` 3.77815125 - 3
` ` ` ` ` ` `t = --------------------
` ` ` ` ` ` ` ` ` `4[log(1.03)]
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` ` ` ` ` ` ` `` ` .7781512504
` ` ` ` ` ` `t = ---------------
` ` ` ` ` ` ` ` ` `.0513488988
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` ` ` ` ` ` ` `` ` .77815125
` ` ` ` ` ` `t = ---------------
` ` ` ` ` ` ` ` ` `.0513488988
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` ` ` ` ` ` `t = 15.15419548 years
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Multiply by 4 to get the number of interest periods
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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 J
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