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 26 hours. How long would it be, to the nearest tenth of an hour, until there are 168 bacteria present?

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  19.5 hours

Step-by-step explanation:

The number in the dish at time t (in hours) can be described by ...

  n(t) = 100·2^(t/26)

We want to find t such that ...

  n(t) = 168

  100·2^(t/26) = 168

  2^(t/26) = 1.68 . . . . . . divide by 100

  (t/26)log(2) = log(1.68) . . . . . take logarithms

  t = 26·log(1.68)/log(2) ≈ 19.459992 . . . . divide by the coefficient of t

  t ≈ 19.5 . . . hours

It would be about 19.5 hours until there are 168 bacteria.

there is no question content here, but you for the free points! : d