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The number of bacteria in a strain is given by $B(t) = 30e^{1.5t}$, where $t$ is the time in hours.

a) How many bacteria are there at time zero?

b) How long will it take the number of bacteria to double?

a) $30e^{1.5t} = 30e^0 = 30$

b) $30e^{1.5t} \ge 2$

$\implies e^{1.5t} = \frac{1}{15}$

$\implies 1.5t = \ln(\frac{1}{15})$

This gives a minus number.

Yuxiao Xie
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dagda1
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    Note the word double here. It's not $2$ but $2B(0)$. – Yuxiao Xie May 03 '16 at 05:10
  • You have a negative number because if at time 0 you have 30, at some time in the past you had 2. Notice you don't want an absolute amount of 2. You want a double the amount of 30. Technically you want to solve for: $\frac {30e^{1.5t}}{30e^0} = 2$. – fleablood May 03 '16 at 06:12
  • In general... B (t)=A$e^{ct} $. Means A is the initial population and $e^{ct} $ is the factor of growth. So you want to simply solve $e^{ct}=2$ which should have the same solution regardless of whatever the initial population was. – fleablood May 03 '16 at 06:17
  • @fleablood great point and very useful for me – dagda1 May 03 '16 at 08:09

1 Answers1

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a) okay : $30~e^{1.5~(0)}=30$

b) You want the time when this is doubled: $30 \, e^{1.5 t}=60~$

$$t = \dfrac{\ln 2}{1.5} $$

Graham Kemp
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