Question

In a lab experiment, 100 bacteria are placed in a petri dish. The conditions are such that the number of bacteria is able to double every 12 hours. How long would it be, to the nearest tenth of an hour, until there are 260 bacteria present?

Answer 1

Answer:

  16.5 hours

Step-by-step explanation:

You know the initial value (100) and the growth factor (2), along with the period associated with that growth factor (12 hours). This lets you write the exponential growth equation as ...

  n = 100×2^(t/12)

For n=260, we can divide by 100 and take the log.

  260 = 100×2^(t/12)

  2.6 = 2^(t/12)

  log(2.6) = (t/12)log(2) . . . . next, divide by the coefficient of t

  12×log(2.6)/log(2) = t ≈ 16.5 . . . . hours

It will be about 16.5 hours until there are 260 bacteria present.

Answer 2

Answer: 16.5 hours

Step-by-step explanation:

this is an example of exponential growth

P = initial * (rate)^x

there are 100 bacteria initially and the rate is 2 since they double, and x would be t/12, because it takes 12 hours to double

so the equation is P = 100 * 2^(t/12)

now substitute 260 in for P and solve for t

260 = 100 * 2^(t/12)

2.6 = 2^(t/12)

t/12 = $log_{2}2.6$

t = 12($log_{2}2.6$)

t = 16.5 hours