Question #b1d2e

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Answer 1 · 2017-01-21T21:53:35

$2, 048, 000$ $2, 048,000$ bacteria

We can say straightaway that

$B (t) = B_{o} \cdot 2^{\frac{t}{0.25}}$ $B(t) = B_o * 2^(t/0.25)$

To check this:

To start with: $B (0) = B_{o} \cdot 2^{0} = B_{o}$ $B(0) = B_o * 2^(0) = B_o$
Then: $B (0.25) = B_{o} \cdot 2^{\frac{0.25}{0.25}} = 2 B_{o}$ $B(0.25) = B_o * 2^(0.25/0.25) = 2 B_o$ so it is doubling :)
And: $B (0.50) = B_{o} \cdot 2^{\frac{0.50}{0.25}} = 4 B_{o}$ $B(0.50) = B_o * 2^(0.50/0.25) = 4 B_o$ , so it has quadrupled. The pattern is established.

Now, if at time 1.25 hours, there are 40,000 bacteria present, we can say that:

$40, 000 = B_{o} \cdot 2^{\frac{1.25}{0.25}} = 5 B_{o} \Rightarrow B_{o} = 8, 000$ $40,000 = B_o * 2^(1.25/0.25) = 5B_o implies B_o = 8,000$

So:

$B (t) = 8, 000 {(2)}^{\frac{t}{0.25}}$ $color(blue)(B(t) = 8,000 (2)^(t/0.25))$

After 2 hours:

$B (2) = 8, 000 \cdot 2^{\frac{2.00}{0.25}}$ $B(2) = 8,000 * 2^(2.00/0.25)$

$= 8, 000 \cdot 2^{8} = 2, 048, 000$ $= 8,000 * 2^(8) = 2, 048,000$