19539 bacteria will be present after 18 hours
Solution:
Initial value of bacteria N = 6000
Value after 4 hours [tex]N_o[/tex] = 7800
The standard exponential equation is given as:
[tex]N=N_{o} E^{-k t}[/tex]
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,
[tex]7800 = 6000E^{-4k}[/tex]
Solving for "k" we get,
[tex]\frac{13}{10}=E^{-4k}[/tex]
Taking "ln" on both sides, we get
[tex]ln\frac{13}{10} = -4k[/tex]
[tex]\frac{ln\frac{13}{10}}{4} = -k[/tex]
On solving for ln, we get k = -0.0656
The equation becomes, [tex]N = 6000E^{0.0656t}[/tex]
Now put "t" = 18,
[tex]N = 6000E^{0.0656 \times 18}\\\\N = 6000 \times 3.256 = 19539[/tex]
Hence the bacteria present after 18 hours is 19539
