Question #d3922

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Answer 1 · 2016-07-04T08:37:29

#t>=log(650/18)/log 1.3. N>=650#.

N is an exponential-growth function and. N increases as t increases.

Equating logarithms and inverting,

#t=log(N/18)/log 1.3#.

So, for #N>=650, t>=log(650/18)/log 1.3=13.6091# (higher rounding))

Answer 2 · 2016-07-04T08:37:48

for # t = 14#

Write an inequality:

#N < 18(1.3)^t " but N < 650"#

#650 < 18(1.3)^t " ÷ 18"#

#650/18 <1.3^t " t is an index , so use logs"#

#log(650/18) < txxlog 1.3#

# t > log(650/18)/log1.3#

#t > 13.6#

So, for t from 14, N will exceed 650.

Test to check ...

If t = 13, then if t = 14....