Answer:
![(log(6400))/(log(4))](https://img.qammunity.org/2020/formulas/mathematics/high-school/7udh51j7m5xbu82ifltsd4ta6fi78spvlk.png)
Explanation:
The initial population of bacteria is 800 and we know that this number is quadrupling every hour.
We're going to find a function in terms of t (time) that gives us the population of bacteria at that time.
Since the population is quadrupling every hour the function in terms of t (where t is expressed in hours) is:
![f(t)=800(4^(t))](https://img.qammunity.org/2020/formulas/mathematics/high-school/348j9k98dzv988hv937xf3nvdv0kxj2mbf.png)
Now we need to find the time when there will be 5,120,000 bacterias. This means the time t when f(t) = 5,120,000
So we have 5,120,000 =
![800(4^(t))](https://img.qammunity.org/2020/formulas/mathematics/high-school/1i75e9qshf5blv4duzyqxjnskwn15jo15m.png)
![5,120,000 = 800(4^(t))\\(5,120,000)/(800) =4^(t) \\6400=4^(t) \\log(6400) =t log(4)\\(log(6400))/(log(4)) =t](https://img.qammunity.org/2020/formulas/mathematics/high-school/l4fe4jlkmcaeaq428xxofvye5xqxvru0iy.png)
Therefore, the time when there will be 5,120,000 bacterias will be:
![(log(6400))/(log(4))](https://img.qammunity.org/2020/formulas/mathematics/high-school/7udh51j7m5xbu82ifltsd4ta6fi78spvlk.png)