1 Answer

Qwerty · Answer 1 · 2024-06-30T10:40:48+0000

Answer:

$p {e}^(20r) = 600$

$p {e}^(30r) = 1000$

${e}^(10r) = (5)/(3)$

r = (1)/(10) ( ln(5) - ln(3) )

$p {e}^(2 ln(5) - 2 ln(3) ) = 600$

(25)/(9) p = 600

p = 216

${e}^{( (1)/(10)( ln(5) - ln(3))t } = 2$

(1)/(10) ( ln(5) - ln(3) )t = ln(2)

$t = (10 ln(2) )/( ln(5) - ln(3) ) = 13.56 \: min$

$p(t )= 600 {e}^{ ((1)/(10) ( ln(5) - ln(3) ))t}$

p(105) = 128104

$600 {e}^{ ((1)/(10) ( ln(5) - ln(3)))t } = 11000$

$t = 56.94 \: min$

- Initial size of culture: 216

- Doubling period: about 13.56 minutes

- Population after 105 minutes: 128,104

- Time when population reaches 11,000: about 56.94 minutes

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