Suppose there are 50 bacteria in pertri dish at a certain time. Six hours later, there are 390 bacteria in the dish.What is the growth constant, k, for this problem?How much bacteria are in the dish after 20 hours?

The growth constant k for this problem is 1.408There are about 46884 bacteria in the dish after 20 hoursStep-by-step explanation:The growth function is [tex]y=a(b)^{x}[/tex] , wherea is the initial valueb is the growth constant factorThe growth equation is [tex]y=a(b)^{x}[/tex]∵ There are 50 bacteria in pertri-dish at a certain time∴ The initial value of bacteria is a = 50∵ The growth constant factor is k∴ [tex]y=50(k)^{x}[/tex], where y is the number of bacteria after x hours∵ Six hours later, there are 390 bacteria in the dish∴ y = 390 and x = 6 - Substitute them in the equation above∴ [tex]390=50(k)^{6}[/tex]- Divide both sides by 50∴ [tex]7.8=(k)^{6}[/tex]- Insert log to both sides∴ [tex]log(7.8)=log(k)^{6}[/tex]∴ ㏒(7.8) = 6 ㏒(k)- Divide both sides by 6∴ 0.1486824 = ㏒(k)- Change ㏒ to base 10∴ [tex]10^{0.1486824}=k[/tex]∴ k = 1.408The growth constant k for this problem is 1.408Substitute the value of k in the equation∴ [tex]y=50(1.408)^{x}[/tex]∵ x = 20∴ [tex]y=50(1.408)^{20}[/tex]∴ y = 46884There are about 46884 bacteria in the dish after 20 hoursLearn more:You can learn more about the equations in brainly.com/question/12363217#LearnwithBrainly