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...