The population of bacteria in a petri dish doubles every 16 h. The population of the bacteria is initially 500 organisms.How long will it take for the population of the bacteria to reach 800?Round your answer to the nearest tenth of an hour.

Accepted Solution

Answer:[tex]10.8\ hours[/tex]Step-by-step explanation:In this problem we have a exponential function of the form[tex]y=a(b)^{\frac{x}{16}}[/tex]wherey ---> is the population of the bacteriax ---> the time in hoursa ---> is the initial value of the populationb ---> is the baser ---> is the rateb=(1+r)where[tex]a=500\ organisms[/tex][tex]r=100\%=100/100=1[/tex][tex]b=(1+r)=1+1=2[/tex]substitute[tex]y=500(2)^{\frac{x}{16}}[/tex]soFor y=800substitute and solve for x[tex]800=500(2)^{\frac{x}{16}}[/tex][tex]1.6=(2)^{\frac{x}{16}}[/tex]Apply log both sides[tex]log(1.6)=log[(2)^{\frac{x}{16}}][/tex][tex]log(1.6)={\frac{x}{16}}log(2)[/tex][tex]x=\frac{log(1.6)}{log(2)}(16)[/tex][tex]x=10.8\ hours[/tex]