Question 1129709
It depends on how frequently the money is compounded.  The most frequent compounding is continuous compunding:  <br>
{{{ F = Pe^(rt) }}}<br>

F = future value
P = present value
r = annual interest rate (expressed as a decimal)
t = number of years<br>

{{{ 2P = P*e^(r(8)) }}}
{{{ ln(2) = 8r }}}
{{{ r = ln(2) / 8 }}} or approx 0.0866   or  {{{ highlight(matrix(1,2, 8.66,"%") ) }}} <br>

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If the money were compounded quarterly (or some other period), use:
{{{ F = P(1 + r/n)^(nt) }}}
n = number of compounding periods per year, other variables are as above.<br>

Continuing with the quarterly example:
{{{ 2P = P(1 + r/4)^(4*8) }}}
{{{ 2 = (1+r/4)^(32) }}}
{{{ ln(2) = 32*ln(1+r/4) }]}
{{{ ln(2)/32 = ln(1+r/4) }}}
{{{ 0.0216608 = ln(1+r/4) }}}
Raise e to each side:
{{{ 1.02190 = 1+r/4 }}}
{{{ r/4 = 0.02190 }}}
{{{ highlight(matrix(1,4, "r =", 8.76, "%", "")) }}}

So not a huge difference (you need a slightly higher rate of return to double your money in 8 years if the interest is not continously compounded).
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Finally, there is this "rule of 70" where you can get an <b>estimate</b> for doubling time (or rate):
   70/r = 8 years
    r = 70/8 = 8.75%    <—<<<  pretty close!