Question A population of bears increased by $50 \%$ in 4 years. If the situation is modeled by an annual growth rate compounded continuously, which formula could be used to find the annual rate according to the exponential growth function? Leave your answer in terms of ln. Note: When entering natural log in your answer, enter lowercase LN as "In". There is no "natural log" button on the Alta keyboard. Provide your answer below: \[ r=\square \] FEEDBACK MORE INSTRUCTION SUBMIT Content attribution
Solution
Step 1 ：The problem is asking for the annual growth rate of a population that increased by 50% in 4 years, modeled by an exponential growth function compounded continuously. The formula for continuous compound interest is \(A = P * e^{rt}\), where \(A\) is the final amount, \(P\) is the principal amount, \(r\) is the rate of interest, and \(t\) is the time. In this case, we know that \(A\) is 150% of \(P\) (since the population increased by 50%), and \(t\) is 4 years. We need to solve for \(r\).
Step 2 ：We can rearrange the formula to solve for \(r\): \(r = \ln(A/P) / t\).
Step 3 ：Let's plug in the values and calculate \(r\).
Step 4 ：Given that \(A = 1.5\), \(P = 1\), and \(t = 4\), we find that \(r = 0.1013662770270411\).
Step 5 ：The annual growth rate, \(r\), is approximately 0.101. Therefore, the formula to find the annual rate according to the exponential growth function is \(r=\boxed{0.101}\)
