Question 3 (3 points) This problem will be two questions (this one on this page and one on the next page - so save the value you got for 3 . ' k !) The George River herd of caribou in Canada was estimated to be about 4,700 in 1954 and grew at an exponential rate to about 472,000 in 1984 . Show how you substitute in the values from the problem to solve the equation to find the growth constant. Then state what the growth constant, k , is for this problem. (round to the nearest thousandth)

To solve for the growth constant \( k \), we can use the exponential growth formula: \[ N(t) = N_0 e^{kt} \] In this case, \( N_0 \) is the initial population (4,700), \( N(t) \) is the population after 30 years (472,000), and \( t \) is 30. Substituting in these values gives us: \[ 472,000 = 4,700 e^{30k} \] Then, divide both sides by 4,700: \[ e^{30k} = \frac{472,000}{4,700} \approx 100.426 \] Now, take the natural logarithm of both sides: \[ 30k = \ln(100.426) \] Calculating that gives: \[ k \approx \frac{\ln(100.426)}{30} \] Finally, rounding to the nearest thousandth, you can find the value of \( k \). --- As you delve into the study of exponential growth, it's fascinating to note that population dynamics not only apply to animal herds like the George River caribou but also to bacteria in a petri dish. In both cases, conditions such as resources and space can greatly influence growth patterns. If you're keen on learning more about population models, books such as "Population Ecology: First Principles" by John H. Vandermeer and Deborah E. G. Vandermeer offer extensive insights. Additionally, looking into online resources and academic papers can provide real-world examples that showcase the importance of understanding growth constants in ecology and conservation efforts.