Asked by Anonymous
Find the amount of time required to double a dollar investment at 7.72%, if the interest is compounded continuously.
4.49 years
8.98 years
17.96 years
I'm not understanding this, I got 3.9 years when I tried to solve this.
4.49 years
8.98 years
17.96 years
I'm not understanding this, I got 3.9 years when I tried to solve this.
Answers
Answered by
MathMate
The basic formula for compounding continuously is:
Lim (1+r/n)^(nt)
as nāā
where t=number of years, and r is the annual interest rate.
which results in the formula:
A(t)=A(0)e^(rt)
For doubling,
A(t)/A(0)=2
Therefore:
e^(0.0772t) = 2
Take natural log on both sides to get:
0.0772t = ln(2)
Solve for t.
Lim (1+r/n)^(nt)
as nāā
where t=number of years, and r is the annual interest rate.
which results in the formula:
A(t)=A(0)e^(rt)
For doubling,
A(t)/A(0)=2
Therefore:
e^(0.0772t) = 2
Take natural log on both sides to get:
0.0772t = ln(2)
Solve for t.
Answered by
Reiny
continuous compounding formula is
amount = principal(e^(it)) , so
2 = 1(e^.0772t)
ln2 = ln (e^.0772t)
.0772 lne = ln2 , but ln e = 1
t = ln2/.0772 = 8.98
how did you get 3.9 ?
amount = principal(e^(it)) , so
2 = 1(e^.0772t)
ln2 = ln (e^.0772t)
.0772 lne = ln2 , but ln e = 1
t = ln2/.0772 = 8.98
how did you get 3.9 ?
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