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 2,500 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 160,000 . 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 . ?
"let the rate be x , then population of the bacteria after each hour can be given as 2500 , 2500 x , 2500 ( x ^ 2 ) , 2500 ( x ^ 3 ) now population at 4 pm = 160,000 thus we have 2500 ( x ^ 3 ) = 160,000 = 64 thus x = 4 therefore population at 3 pm = 2500 ( 16 ) = 40,000 answer : b"
multiply(multiply(power(divide(multiply(multiply(2, 4), 2,500), 2,500), const_0_33), 2,500), power(divide(multiply(multiply(2, 4), 2,500), 2,500), const_0_33))