+118609 $$\\T_n=ar^{n-1}\\\\
3*2^{n}>100\\\\
2^{n}>100/3\\\\
log2^{n}>log(100/3)\\\\
nlog2>log(100/3)\\\\
n>log(100/3)\div log2\qquad\\\\$$
$${\frac{{log}_{10}\left({\frac{{\mathtt{100}}}{{\mathtt{3}}}}\right)}{{log}_{10}\left({\mathtt{2}}\right)}} = {\mathtt{5.058\: \!893\: \!689\: \!053\: \!568\: \!6}}$$
n has to be greater than this so 6 days.
-------------------------------------------------
I failed to mention a couple of things here in this GP sequence a=3 and r=2
but the number wanted is the n+1 because it is the END of that day. So I changed the sqeuence around a little and made 3 be the 'zero' term and then I was looking for the n'th term.
I hope I didn't s***w it up That would not be good 
+118609 $$\\T_n=ar^{n-1}\\\\
3*2^{n}>100\\\\
2^{n}>100/3\\\\
log2^{n}>log(100/3)\\\\
nlog2>log(100/3)\\\\
n>log(100/3)\div log2\qquad\\\\$$
$${\frac{{log}_{10}\left({\frac{{\mathtt{100}}}{{\mathtt{3}}}}\right)}{{log}_{10}\left({\mathtt{2}}\right)}} = {\mathtt{5.058\: \!893\: \!689\: \!053\: \!568\: \!6}}$$
n has to be greater than this so 6 days.
-------------------------------------------------
I failed to mention a couple of things here in this GP sequence a=3 and r=2
but the number wanted is the n+1 because it is the END of that day. So I changed the sqeuence around a little and made 3 be the 'zero' term and then I was looking for the n'th term.
I hope I didn't s***w it up That would not be good
