Question

A culture started with 6,000 bacteria. After 4

Answer 1

19539 bacteria will be present after 18 hours

Solution:

Initial value of bacteria N = 6000

Value after 4 hours $N_o$ = 7800

The standard exponential equation is given as:

$N=N_{o} E^{-k t}$

where

N is amount after time t

No is the initial amount

k is the constant rate of growth

t is time

Plugging in the values in formula we get,

$7800 = 6000E^{-4k}$

Solving for "k" we get,

$\frac{13}{10}=E^{-4k}$

Taking "ln" on both sides, we get

$ln\frac{13}{10} = -4k$

$\frac{ln\frac{13}{10}}{4} = -k$

On solving for ln, we get k = -0.0656

The equation becomes, $N = 6000E^{0.0656t}$

Now put "t" = 18,

$N = 6000E^{0.0656 \times 18}\\\\N = 6000 \times 3.256 = 19539$

Hence the bacteria present after 18 hours is 19539