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?

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

[tex]Bacterias(55hours)=C*r^{{\frac{55-t}{t}}} \\Bacterias (66hours) = C*r^{\frac{66-t}{t}}}[/tex]

If you divide both you can get the growth rate:

[tex]\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}}[/tex]

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

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

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?

Respuesta :

Otras preguntas

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?