
#19
Apr310, 12:10 PM

P: 42

does the equation become
just for the subtitution [itex] \frac{1}{ln(1000P)} [/itex] 



#20
Apr310, 01:09 PM

HW Helper
P: 1,495

No. It appears that you do not know how to integrate by substitution. Looking at the problem you have presented you ought to have followed a course at some point that taught you how to do it. I suggest reviewing the basics of integration before continuing with this problem.




#21
Apr310, 11:17 PM

P: 42

[itex] \frac{1}{5} lnP  \frac{1}{5} ln(1000P)= t+C [/itex]
[itex] lnP  ln(1000P) = 5t+5c [/itex] using log rules [itex] ln \frac{P}{1000P} = 5t+5c [/itex] multiplying by e [itex] \frac{P}{1000P}=Ae^{5t} , A=e^{5c} [/itex] [itex] P = (1000P)(Ae^{5t}) [/itex] multiplying it out [itex] P=1000Ae^{5t}  PAe^{5t} [/itex] [itex] P + PAe^{5t} = 1000Ae^{5t} [/itex] [itex] Ae^{5t}= 1000Ae^{5t}/P [/itex] [itex] P = 1000 [/itex] but how do i get P(t) 



#22
Apr410, 03:34 AM

P: 42

turns out i just had to use rearranging to find my P(t)



Register to reply 
Related Discussions  
Population growth help  Calculus & Beyond Homework  3  
Population Growth  Differential Equations  8  
Population Growth  Calculus & Beyond Homework  6  
population growth  Introductory Physics Homework  3  
Population Growth  General Math  2 