Nahhhhhhh i don’t man sorry though bro this is though like buttocks on a stick XDXDXD
Since, population of species A is represented by : 
Let us find the population of species A, at the end of week 1:
i.e., x = 1
i.e., 
i.e., 
i.e., 
Also, since population of species B is represented by : 
Let us find the population of species B, at the end of week 1:
i.e., x = 1
i.e., 
i.e., 
i.e., 
Thus, at the end of 1 week, species A and species B will have the same population.
Hence, option D is correct.

Step-by-step explanation:
