Given the function p(t) and the initial condition, we have the following:
[tex]\begin{gathered} p(t)=p_0\cdot1.42^t \\ p(0)=1000 \\ \Rightarrow p(0)=p_0\cdot1.42^0=1000 \\ \Rightarrow p_0\cdot1=1000 \\ p_0=1000 \end{gathered}[/tex]Therefore, the function p(t) is defined like this:
[tex]p(t)=1000\cdot1.42^t[/tex]Now, since we want to know the time it will take the population to exceed 10,000 cells, we have to solve for t using this information like this:
[tex]\begin{gathered} p(t)=1000\cdot1.42^t=10000 \\ \Rightarrow1.42^t=\frac{10000}{1000}=10 \\ \Rightarrow1.42^t=10 \end{gathered}[/tex]Applying natural logarithm in both sides of the equation we get:
[tex]\begin{gathered} 1.42^t=10 \\ \Rightarrow\ln (1.42^t)=\ln (10) \\ \Rightarrow t\cdot\ln (1.42)=\ln (10) \\ \Rightarrow t=\frac{ln(10)}{\ln (1.42)}=6.56 \end{gathered}[/tex]Therefore, it will take the population 6.56 hours to exceed 10,000 cells
