Al P. answered • 02/26/18

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For continuous compounding you need to let n—>∞

which turns A=P(1+r/n)

^{nt }into**A=Pe**^{rt}The amount of principal does not affect the doubling time, so I'll just leave it as P until we divide it out:

2P = Pe

^{0.05t} 2 = e

^{0.05t }(divided both sides by P)ln(2) = 0.05t

Solve for t.

(I got approx 13.863 years)

If your teacher is insistent on using A=P(1+r/n)

^{nt}then a close approximation can be obtained by using a large n (say n=365 for daily compounding, your answer will be pretty close).