Problem: david works at a science lab that conducts experiments on bacteria . the population of the bacteria multiplies at a constant rate , and his job is to notate the population of a certain group of bacteria each hour . at 1 p . m . on a certain day , he noted that the population was 600 and then he left the lab . he returned in time to take a reading at 4 p . m . , by which point the population had grown to 4,800 . now he has to fill in the missing data for 2 p . m . and 3 p . m . what was the population at 3 p . m . ?
Rationale: let the rate be x , then population of the bacteria after each hour can be given as 600,600 x , 600 ( x ^ 2 ) , 600 ( x ^ 3 ) now population at 4 pm = 4800 thus we have 600 ( x ^ 3 ) = 4800 = 8 thus x = 2 therefore population at 3 pm = 600 ( 4 ) = 2400 answer : a
annotated_formula: multiply(multiply(power(divide(multiply(multiply(2, 4), 600), 600), const_0_33), 600), power(divide(multiply(multiply(2, 4), 600), 600), const_0_33))