Bacterial population doubles every hour. If there are 1200 bacteria now, when was 300 bacteria present?

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Solution

The correct option is A 2 hours ago
Given,
A = 1200
P = 300
r = 100% per hour [population doubles after every hour]

Let the number of hours for the population of bacteria to become 1200 be n.

$A = P (1 + \frac{100}{100})^{n}$
$1200 = 300 (1 + 1)^{n}$
$4 = (2)^{n}$
$n = 2 h o u r s$

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Bacterial population doubles every hour. If there are 1200 bacteria now, when was 300 bacteria present?

The correct option is A 2 hours agoGiven, A = 1200 P = 300 r = 100% per hour [population doubles after every hour] Let the number of hours for the population of bacteria to become 1200 be n. A=P(1+100100)n 1200=300(1+1)n 4=(2)n n=2 hours

The correct option is A 2 hours ago
Given,
A = 1200
P = 300
r = 100% per hour [population doubles after every hour]

Let the number of hours for the population of bacteria to become 1200 be n.

$A = P (1 + \frac{100}{100})^{n}$
$1200 = 300 (1 + 1)^{n}$
$4 = (2)^{n}$
$n = 2 h o u r s$