How Many Years Will It Take These Accounts To Reach The Same Amount? A $1000 Investment At A Rate Of 3% Compound Monthly And $1100 Investment At A Rate Of 3.02% Compounded Annually.
Answer:
The first account will yield after n years
f(n) = 1000(1+3%/12)^(12n)
The second account will concede after n years
s(n) = 1100(1+3.02%)^n
When the values of these accounts are equal, the log of their values will be equal
Log[f(n)] = Log[s(n)]
Log[1000] + 12nLog[(1.0025)] = Log[1100]+nLog[(1.0302)]
Solving for n, we get
n(12Log[1.0025] - Log[1.0302]) = Log[1100] - Log[1000]
n = Log[1.1]/(12Log[1.0025] - Log[1.0302])
n = .041393/(12*.0010844 - .012922) (note the small difference involved here. Greater precision is required of these numbers than is given here)
n = .041393/.000091030
n ≈ 454.71
It will take about 455 years for the first justification to match the second in total value.
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