1. The given situation can be described in the following inequality:
[tex]216^{(t+18)}<36^{(2t+8)}[/tex]
2. To solve the prevous inequality, proceedas follow:
write 216 as 6^3 and 36 as 6^2, then, apply log_6 both sides:
[tex]\begin{gathered} 6^{3(t+18)}<6^{2(2t+8)} \\ \log _66^{3(t+18)}<\log _66^{2(2t+8)} \\ 3(t+18)<2(2t+8) \end{gathered}[/tex]
then, solve for t, as follow:
[tex]\begin{gathered} 3t+54<4t+16 \\ 3t-4t<16-54 \\ -t<-38 \\ t>38 \end{gathered}[/tex]
Hence, from t = 38 hours the number of type A bacteria will be less than the type B bacteria
3. In a number line, you obtain:
