In a study by the texas a & m university, found there were 23,000 prairie dogs in texas in 2000. by 2010, the number has increased to 40,000. the exponential growth function d=2.3e^kt describes the prairie dog population, d in ten thousands, t years after 2000.
a) solve for k
b)by this model how many prairie dogs existed last year?
c)in which year will the population double?

The population function is given as .

(A) " alt="D=2.3e^{kt}\\\frac{D}{2.3}=e^{kt}\\\ln\frac{D}{2.3}=\ln e^{kt}\\\ln\frac{D}{2.3}=kt\\k=\frac{\ln\frac{D}{2.3}}{t}\\k = \frac{\ln\frac{4}{2.3}}{10}=0.0553" />" />

The value of k is 0.0553.

(I used D=4 for t=10 to determine the value of k.)

(B) With the above, we have a fully determined growth function:

and can answer this question now. Last year was 2018, so t=18:

" alt="D(18)=2.3e^{0.0553\cdot 18 }=6.22" />" />

Last year (2018), there existed approximately 62200 prairie dogs.

(C) We are looking for t* for which D(t*) will be twice the population as of last year (Note: this is my interpretation of the vague question wording - it is not 100% clear with respect to which population we are considering the doubling):

The doubling will occur approximately in 2030 (mid-year).

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step-by-step explanation:

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      2

  2 is to go somewhat to the top right of a because it is squared    

v= a h/3