Okay i have no idea how to start this question.... so if any one can help out please...
In a certain forest there are two species of moth, grey and black. presently there are twice as many grey as black, but the intrinsic growth rate of the black moth is 4% per month, while that of the grey is only 3% per month. in how many months will there be twice as many black as grey
Answer: (11.5 a)
In a certain forest there are two species of moth, grey and black. presently there are twice as many grey as black, but the intrinsic growth rate of the black moth is 4% per month, while that of the grey is only 3% per month. in how many months will there be twice as many black as grey
Answer: (11.5 a)
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G(t) = Number of grey moths
G(t) = G(0)e^(.03t)
B(t) = Number of black moths
B(t) = B(0)e^(.04t)
t = number of months since t = 0
G(0) = 2B(0) : currently twice as many Grey as Black moths
Set B(t) = 2G(t) and solve for t to find when there will be twice as many Black as Grey:
B(0)e^(.04t) = 2G(0)e^(.03t)
B(0)e^(.04t) = 2[2B(0)]e^(.03t)
e^(.04t) = 4e^(.03t)
e^(.01t) = 4
.01t = ln4
t = ln4/.01 = 138.6 months (11.55 years)
G(t) = G(0)e^(.03t)
B(t) = Number of black moths
B(t) = B(0)e^(.04t)
t = number of months since t = 0
G(0) = 2B(0) : currently twice as many Grey as Black moths
Set B(t) = 2G(t) and solve for t to find when there will be twice as many Black as Grey:
B(0)e^(.04t) = 2G(0)e^(.03t)
B(0)e^(.04t) = 2[2B(0)]e^(.03t)
e^(.04t) = 4e^(.03t)
e^(.01t) = 4
.01t = ln4
t = ln4/.01 = 138.6 months (11.55 years)