alevel maths differential equations question

how would you do this??

rate of change of a population is given by the formula dP/dt = 2P
time is t
given initial population in P0 is 1000 find the expression for P in terms of t

This is a first order differential equation of the type that can be solved by separation of variables. Can I suggest that you check "first order differential equation" and "separation of variables" in you textbook or notes? Then have a go and post your working if stuck.

i know the method but i am stuck on the last step : ln |2p| = t + ln |2000|

First you need to check that left hand side, as the integral of (1 / 2P) is not ln(2P).

But then, once you have the equation in the form ln(something) = kt + c, you can take the exponential of both sides to give you:

something = e^(kt + c) = e^(kt) x e^(c) = Ae^(kt)

Don't mind me butting in, but I got ln(2P) = 2t +ln(2000) [Multiplied both sides by 2 to get rid of the halves] ... and I took exponentials to make 2P the subject....
2P = e^(2t +ln(2000)).
So P = 1/2 e^(2t+ln(2000))

This seems a little clunky... Although it is technically an expression for P in terms of t.... any simplifying I could do?

Don't mind me butting in, but I got ln(2P) = 2t +ln(2000) [Multiplied both sides by 2 to get rid of the halves] ... and I took exponentials to make 2P the subject....
2P = e^(2t +ln(2000)).
So P = 1/2 e^(2t+ln(2000))

This seems a little clunky... Although it is technically an expression for P in terms of t.... any simplifying I could do?

You could note that e^(2t + ln(2000)) = e^(2t) x e^(ln(2000)) = Ae^(2t)

How come the answer is written as P= e^(2t+20√10)

If you put t = 0 in that formula, I don't think you'll get P0 = 1000.

i got P=1000e^2t ?

How come the answer is written as P= e^(2t+20√10)