Question 1179797
more than one way to do this



{{{800*(1+r)^2=1280}}}

{{{1+r=sqrt(1280/800)}}}

{{{r=sqrt(128/80)-1}}}

{{{r=sqrt(8/5)-1}}}
and then you can say  {{{f(t)=800(1+r)^t}}}.  
That would be for t in HOURS.  You want the appropriate adjustment for DAYS.




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{{{f(t)=800(1.03)^t}}}
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For 2 days, t is 48.