Question

A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases exponentially with time. At the end of 55 hours there are 192192​,000 bacteria. At the end of 66 hours there are 384384​,000. How many bacteria were present​ initially?

Answer 1

Answer:

Initial bacterias = 6006000

Altought I believe is safe to assume that the values were 192,000 and 384,000 instead of 192,192,000 and 384,384,000, in that case the initial bacterias is 6000

Step-by-step explanation:

A exponential growth follows this formula:

Bacterias  = C*rⁿ

C the initial amount

r the growth rate

n the number of time intervals

Bacterias (55 hours) = 192,192,000

Bacterias (66 hours) = 384,384,000

$Bacterias(55hours)=C*r^{{\frac{55-t}{t}}} \\Bacterias (66hours) = C*r^{\frac{66-t}{t}}}$

If you divide both you can get the growth rate:

$\frac{Bacterias (66hours)}{Bacterias(55hours)}=\frac{C*r^{\frac{66-t}{t}}}{C*r^{{\frac{55-t}{t}}}} \\\frac{384,384,000}{192,192,000} =r^{\frac{66-t}{t} -\frac{55-t}{t} } \\2 =r^{\frac{11}{t}}$

So with that r = 2 and each time interval correspond to 11 years

Then replacing in one you can get the initial amount of C

$Bacterias (55hours)=C*2^{\frac{55-11}{11} } 192,192,000 = C*32\\C= 6006000$