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]
where
y ---> is the population of the bacteria
x ---> the time in hours
a ---> is the initial value of the population
b ---> is the base
r ---> is the rate
b=(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]
so
For y=800
substitute 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]
