A sample of 137 million bacteria is observed to grow to 17 m

A sample of 1.37 million bacteria is observed to grow to 1.7 million after 1 hour. That sample is set aside for an unknown time, and then mixed with some number of a second species of bacteria. The two species do not interact. At the time of mixing, the are a total of 7.6 million bacteria; two hours later 10.3 million, and 4 hours later, 14.1 million bacteria. How many individuals of the second species are present in the mixture after 8 hours? Suggestions: if the growth rates of the first species are k_1, k_2 then, after setting u_1 = exp(2K_1) and u = exp(2k_2), show that you end up solving expiations like y_1 = A + B, y_2 = Au_1 + Bu_2, and Y_3 = A(u_1)^2 + B(U_2)^2, where y_1 = 7.6, Y_2 = 10.3, and y_3 = 14.1. Solve those equations algebraically; put the numbers in after you solve them.

Solution

Please note that the given problem is related to Calculus, and only basic algebraic knowledge is required. Kindly post the question in appropriate medium. Thank you, and wish you all the best.