Question 569105
If {{{ t }}} is the time gained or lost the 1st year,
After 30 yrs:
{{{ (1/2)*t }}}
Next 30 yrs:
{{{ (1/4)*t }}}
Next 30 yrs:
{{{ (1/8)*t }}}
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So, at the end of {{{ n*30 }}} yrs,
The accuracy is {{{  (1/(2^n)) * t }}}
If {{{ n*30 = 1700 - 1350 }}}
{{{ n = 350 / 30 }}}
{{{ n = 11.667 }}}
{{{ t = 30 }}} min/day
{{{ (1/( 2^11.667))*30 }}} min/day
You can use log to solve this. Call it {{{ a }}}
for accuracy
{{{ a = (1/( 2^11.667))*30 }}}
{{{ log(a) = -11.667*log(2) + log(30) }}}
When you find {{{ log(a) }}}, then
{{{ a = 10^((log(a)) ) }}}
( I don't have calculator handy )