Question 438977
B = A(1+i)^N + (P/i)[(1+i)^N − 1]

Solve for "N":
A(1+i)^N + (P/i)(1+i)^N = B+1
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[A + (P/i)](1+i)^N = B+1
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(1+i)^N = (B+1)/[A+(P/i)]
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Take the log of both sides:
N*log(1+i) = log[(B+1)/((Ai+P)/i)]
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N = {log[(B+1)/((Ai+P)/i)]}/log(1+i)
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Cheers,
Stan H.
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