Compound countinuously, it would only take blank years.
A=Pe^rt
A=6000
P=3000`
e=e
r=0.1
t=?
6000=300e^0.1*t
3 answers
I meant to put 3000 for P
A=Pe^rt
6000/3000=3000/3000e(0.1t)
2=e^(0.1t)
In 2=In (e^(0.1t)
In 2= 0.1t(In e)
0.1t=In(2)
t=In(2)/0.1
=0.693147181/0.1
=6.931471806=7 years.
6000/3000=3000/3000e(0.1t)
2=e^(0.1t)
In 2=In (e^(0.1t)
In 2= 0.1t(In e)
0.1t=In(2)
t=In(2)/0.1
=0.693147181/0.1
=6.931471806=7 years.
I agree, assuming the question was, "how long would it take for money to double at 10% compounded continuously?"
( In the good ol' days we had something called the rule of 72. Money doubles approximately if you multiply the time by the rate and you get 72.
So 72/10 = 7.2 years, which is close to the 7 years of your answer)
( In the good ol' days we had something called the rule of 72. Money doubles approximately if you multiply the time by the rate and you get 72.
So 72/10 = 7.2 years, which is close to the 7 years of your answer)
2 answers