Question #d3922

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Answer 1 · 2016-07-04T08:37:29

$t \geq \frac{log (\frac{650}{18})}{log 1.3} . N \geq 650$ $t>=log(650/18)/log 1.3. N>=650$ .

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

Equating logarithms and inverting,

$t = \frac{log (\frac{N}{18})}{log 1.3}$ $t=log(N/18)/log 1.3$ .

So, for $N \geq 650, t \geq \frac{log (\frac{650}{18})}{log 1.3} = 13.6091$ $N>=650, t>=log(650/18)/log 1.3=13.6091$ (higher rounding))

Answer 2 · 2016-07-04T08:37:48

for $t = 14$ $t = 14$

Write an inequality:

$N < 18 {(1.3)}^{t} but N < 650$ $N < 18(1.3)^t " but N < 650"$

$650 < 18 {(1.3)}^{t} \div 18$ $650 < 18(1.3)^t " ÷ 18"$

$\frac{650}{18} < {1.3}^{t} t is an index , so use logs$ $650/18 <1.3^t " t is an index , so use logs"$

$log (\frac{650}{18}) < t \times log 1.3$ $log(650/18) < txxlog 1.3$

$t > \frac{log (\frac{650}{18})}{log 1.3}$ $t > log(650/18)/log1.3$

$t > 13.6$ $t > 13.6$

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

Test to check ...

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