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# A bacteria population is initially 320. After 45 minutes, they've grown in number to 700. What is the doubling time for this population? ?

39 minutes 41 seconds

39 minutes 31 seconds

39 minutes 21 seconds

39 minutes 51 seconds

### 3 Answers

- KrishnamurthyLv 74 months ago
A bacteria population is initially 320. After 45 minutes,

they've grown in number to 700.

What is the doubling time for this population?

Bacteria population's doubling time for this population is

39 minutes 51 seconds

Explanation:

Let the doubling time of bacteria population be t.

This means that after t minutes, it will be doubled and after nt minutes

it will grow 2^n times.

Hence in the instant case

700 /320 = 2^(45/t) or 21.875 = 2^(45/t)

or log 2 ( 2.1875 ) = 45/t

or log 2.1875 log 2 = 45/t

or *0.33995/0.30103 = 45/t #

i.e. t = 45⋅0.30103

0.33995 = 39.05 minutes

Hence bacteria population's doubling time for this population is

39.05 minutes

39 minutes 51 seconds

- Jeff AaronLv 74 months ago
Assuming exponential growth:

y = ab^x

a = initial population = 320

y = 320 * b^x

When x = 45, y = 700:

700 = 320 * b^45

b^45 = 700/320 = 35/16 = 2.1875

b = 2.1875^(1/45)

y = 320 * 2.1875^(x/45)

After the doubling time, y is 2 * 320 = 640:

320 * 2.1875^(x/45) = 640

2.1875^(x/45) = 640 /320 = 2

x/45 = log[2.1875](2)

x = 45*log[2.1875](2) =~ 39.848292522575846712897275462287

Rounding off, that's 39 minutes and 51 seconds.

- Wayne DeguManLv 74 months ago
B(t) = 320 x e^kt

When t = 45, B(t) = 700

so, 700 = 320 x e^45k

=> e^45k = 35/16

so, 45k = ln(35/16)

i.e. k = (1/45)ln(35/16)

Then, B(t) = 320 x e^[(t/45)ln(35/16)]

Hence, B(t) = 320 x (35/16)^(t/45)

Doubling occurs when B(t) = 640

i.e. when (35/16)^(t/45) = 2

so, (t/45)ln(35/16) = ln2

Hence, t = 45ln2/ln(35/16) => 39 minutes and 51 seconds

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