What interest rate would be necessary to double a 100 investment in 11 years?

Suppose the interest rate is r%

then 2 = (1+r/100)11

so that ln(2) = 11*ln(1+r/100)

Then ln(1+r/100) = ln(2) /11

so 1 +r/100 = exp[ln(2)/11]

so r/100 = exp[ln(2)/11] - 1

r = 100*{exp[ln(2)/11] - 1} = 6.504 % approx.