exponential growth - relative growth rate

[PepprRN]

I have searched the forum for a problem like this and can't find one using the formula's I have been given.
I have an exponential problem I need help with: Grey Squirrels introduced 30 years ago. The population doubles after 6 years. Now the population is 100,000.

A. What was the initial population?

I used n(t)=n<sub>0*2^t/a

100000=N<sub>0*2^30/6
100000=n<sub>0*32
3125=N<sub>0

Here's where I get hung up....

B. What is the population 10 years from now...

I set it up with N(t)= n<sub>0 e ^rt

N(10)= 100000e^(6)(10)

I can't get it to come out right....I am doing something wrong in solving this!!

This is what I am doing:
N(10)=100000e^(6)(10)
n(10)/100000=e^60

and there is where i get stuck....the numbers are too small(big) to work with.....Help!!

 

[Deleted member 4993]

I have searched the forum for a problem like this and can't find one using the formula's I have been given.
I have an exponential problem I need help with: Grey Squirrels introduced 30 years ago. The population doubles after 6 years. Now the population is 100,000.

A. What was the initial population?

I used n(t)=n<sub>0*2^t/a

100000=N<sub>0*2^30/6
100000=n<sub>0*32
3125=N<sub>0

Here's where I get hung up....

B. What is the population 10 years from now...

I set it up with N(t)= n<sub>0 e ^rt ....why ??? Your guiding equation (for squirrel population) is:\(\displaystyle \ n(t) \ =\ n_0*2^{\frac{t}{a}}\)

N(10)= 100000e^(6)(10)

I can't get it to come out right....I am doing something wrong in solving this!!

This is what I am doing:
N(10)=100000e^(6)(10)
n(10)/100000=e^60

and there is where i get stuck....the numbers are too small(big) to work with.....Help!!

.

 

[HallsofIvy]

"r" is NOT 6. If the population doubles in 6 years then 2N= Ne^{6r} so that 6r= ln(2) and r= ln(2)/6.

Of course, then N(t)= N(0)e^([ln(2)/6]t]= N(0)e^ln(2t/6)= N(0)(2^(t/6)) as you used before.