Arthur D. answered 08/15/15
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1700*(0.7)^y=100 or less where y=# of years it takes
(0.7)^y=100/1700
(0.7)^y=1/17
log([0.7]^y)=log(1/17)
y*log(0.7)=log(1/17)
y=log(1/17)/log(0.7)
y=-1.2304489/-0.1549019
y=7.9≈8 years
1700*0.7=1190 1
1190*0.7=833 2
833*0.7=583.10 3
583.10*0.7=408.17 4
408.17*0.7=285.72 5
285.72*0.7=200 6
200*0.7=140 7
140*0.7=$98 8