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If you invest a sum of money in an account earning 16% annual interest compounded continuously, how long will it take for your money to double?
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\[A= Pe^{rt}\] A= final amount = 2P P= initial amount r= rate = 16% = 0.16 t = time \[2P = Pe^{rt}\] \[2= e^{0.16t}\] take ln of both sides, remember that ln(e) = 1 \[\ln(2) = \ln(e^{0.16t})\] \[\ln(2) = 1(0.16t)\] \[t= \frac{ \ln(2) }{ 0.16}\] t=? You can solve this :)
Thank you!!!!!!! I had most of those steps I just didn't know the natural log part!
You're welcomed :) Yeah that's part a little tricky
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