Q: How can I solve a population growth problem using the general equation?

[ksle82]

im trying to solve a population growth problem. not sure if I am using the right eqn. Please check:

Q: if the population doubles in two years, how long does it take to triple?

Solution:

general equation for population growth: N=No*e^(rt)
1) find unknown constant "r" from given
2No=No*e^(rt)
from equation, r= ln(2)/2
2) find "t" for population to triple
3No=No*e^(rt)
solving for t: t=2ln(3)/ln(2) ~ 3.170 years

 

[neurocomp2003]

If that's the equation your given then yes you are using the right equation. Just verify that t is indeed measured in years rather than some other time scale. Also verify that they imply triple to mean 3No not 3(2No). Even thoughy you used a calculator i think you can still simplify ln(3)/ln(2)

 

[ksle82]

If that's the equation your given then yes you are using the right equation.

thats the problem I am not sure I am using the right equation.

 

[HallsofIvy]

Yes, that's a perfectly valid formula.

However, you could also use
[tex]P(t)= P_0 2^{\frac{t}{2}}[/tex]
(every two years, t/2 is an integer, so we have multiplies by 2 t/2 times.)
then
[tex]P(t)= 3P_0= P_0 2^{\frac{T}{2}[/tex]
[tex]2^{\frac{T}{2}}= 3[/tex]
[tex]\left(\frac{T}{2}\right) log(2)= log(3)[/tex]
[tex]T= \frac{2 log(3)}{ log(2)}[/tex]
as you have.

 

[neurocomp2003]

also note that your equation when substituted with your given value of "r" simplies to the equation posted by HallsOfIvy

 

[HallsofIvy]

Exactly. All "exponentials" are interchangable. That's why you only need log base 10 and log base e on your calculator.
[tex]2^{\frac{t}{2}}= e^{ln(2^{\frac{t}{2}}= e^\frac{t}{2}ln(2)[/tex]
which is [itex]e^{rt}[/itex] with r= ln(2)/2.