Population grows according to the equation , where k is a constant and t is measured in years. If the population doubles every 10 years, then the value of k is
(a) 0.069 (b) 0.200 (c) 0.301 (d) 3.322 (e) 5.000
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Population grows according to the equation , where k is a constant and t is measured in years. If the population doubles every 10 years, then the value of k is
(a) 0.069 (b) 0.200 (c) 0.301 (d) 3.322 (e) 5.000
Hello,Quote:
Originally Posted by frozenflames
the growth of this population can be described by the equation:
where p is the actual amount,is the starting value.
Put in the values you know:
solve for k:
now you have to use a caculator or a logarthmic table.
You'll get
that means that none of the given results seem to be right.
Greetings
EB
Greetings EB:
I solved for k in the equation P(t) = P_0 e^kt which yields k = 0.1*ln(2) approx= 0.069. As it happens, 0.069 is indeed one of the indicated choices.
Enjoy the day,
Rich B.
Hello, RichB.Quote:
Originally Posted by Rich B.
of course you are right - BUT. when frozenflames referred to the equation I was not aware, that the equation was meant which you used. It all depends on the base you choose for solving this problem. I have choosen for 10 and thats why I came up with a "wrong" result.
Have a nice day too.
EB
In fact it does not matter (in principle) what you take forQuote:
Originally Posted by earboth
(as long as it is positive :D ) in a growth equation of the form:
for:
Here there is a good case for using 2 since we are discussing doubling
times, so here it might be natural to use:
RonL
Hi Ron:
Right you are. I was just fishing for a base that would yield a parameter, k, that appears in the multiple choice list.
Enjoy.
Rich