This is a solved problem. Deposit $1000 @ 8% compounded continuously. How many years before it doubles. Solved: Use A=Pe^.08 , so 2000=1000e^.08n then 2=e^.08n then take natural log (1n) of both sides. end up with 1n 2 = .08n
so that n=
1n 2/.08 approx= .6931/.08 approx=8.66 so 8yr8mo
My question is where did the .6931 come from.
5 answers
.6931 = ln(2)
2 = e^(.08n)
ln 2 .08 n
but ln 2 = .6931471806
ln 2 .08 n
but ln 2 = .6931471806
I'm so confused. How did you figure it out.
Terri,
Let me start at the top.
1000 at .08
A = 1000 e^ .08 t if continuous compounding
double is 2000
so
2000 = 1000 e^.08 t
or
2 = e^.08 t
now take the natural log of both sides
ln 2 = ln [e^.08t)
or
ln 2 = .08 t because ln (e^anything) = anything
now if only we knew ln 2 , we would have this solved.
But we do. Do ln 2 on your calculator or look it up in a table of natural logs and you will find that
ln 2 = .69314 etc
then
.69314 = .08 t
Let me start at the top.
1000 at .08
A = 1000 e^ .08 t if continuous compounding
double is 2000
so
2000 = 1000 e^.08 t
or
2 = e^.08 t
now take the natural log of both sides
ln 2 = ln [e^.08t)
or
ln 2 = .08 t because ln (e^anything) = anything
now if only we knew ln 2 , we would have this solved.
But we do. Do ln 2 on your calculator or look it up in a table of natural logs and you will find that
ln 2 = .69314 etc
then
.69314 = .08 t
Great. Thanks so much