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@VeritasVosLiberabit
\[A(t)=Pe ^{rt}\] \[100=50e ^{.05t}\] \[\ln(\frac{ 100 }{ 50 })=\ln(e ^{.05t})\] \[\ln(2)=.05t\] \[\frac{ \ln(2) }{ .05 }=t\] t=13.9 years
Does this make sense? P is the starting quantity and A(t) is the final. r is the rate and t is time
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Yes it does o:
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