# proportional growth - interest rate

• Dec 14th 2009, 03:42 PM
DBA
proportional growth - interest rate
How long will it take an investment to double in value if the interest rate is 6% compounded continuously?

I did:

A = A(0) * e^rt

r=0.06
Set A(0)=1000
Set A=2000

2000 = 1000 * e^0.06 * t
ln2 = ln e^0.06 * t
ln2 = 0.06*t
t= ln2/0.06 = 11.55 years

What is the equivalent annual interest rate?

I did
A(t)= A(0)*(1+ r/n)^nt

2000 = 1000 (1 + r/1)^ 1*11.55
2 = (1 + r/1)^ 1*11.55

Here I stuck. Can someone let me know if it is right what I did and how to get the interest rate?

I thought I could do
ln2 = ln(1 + r/1)^ 1*11.55
ln2 = 11.55 * ln(1 + r/1)

but the I stuck with that...
• Dec 14th 2009, 09:36 PM
nehme007
Compounded continuously, your investment will grow according to:
\$\displaystyle A(t) = A(0) e^{rt} \$
Compounded annually, your investment will grow according to:
\$\displaystyle A(t) = A(0) (1+r)^t \$,
where t is measured in years. The former is really just \$\displaystyle A(t) = A(0) (e^r)^t \$, which is in the same form as the latter but with \$\displaystyle e^r \$ in place of \$\displaystyle 1+r\$. So the equivalent rate if compounded annually is \$\displaystyle e^r - 1\$, which comes out to something like 6.185%.

I think you get the same thing if you solve the equation you wrote down. You don't need to take logs to solve it, however:
\$\displaystyle 2 = (1 + r)^{11.55} \$
\$\displaystyle 2^{1/11.55} = 1 + r \$