Answer:
t=11,11 days
Step-by-step explanation:
F=frogs poblation, t=time, be the variables dF/dt = KF, dF/F=Kdt, integrating [tex]\int\limits^ {} \, dF/F =K\int\limits^ {} \, dt[/tex]⇒ LnF=Kt+c,[tex]F=ce^{Kt}[/tex]; Knowing t=0, F=17 and t=6 F=51 (tripling every 6 days (17*3)), [tex]F=ce^{K0} = F=c=17; F=17e^{6K} =51[/tex]⇒[tex]e^{6K} =51/17; K6=ln\frac{51}{17} ; K=ln\frac{51}{17}/6=0.183[/tex], so [tex]F=17e^{0,183t}[/tex], now if F=130, t=? we have:
[tex]130=17e^{0.183t} =e^{0.183t} =130/17; 0.183t=ln(130/17); t=ln(130/17)/0.183 = 11,11[/tex]
