Colby and jaquan are growing bacteria in an experiment in a laboratory. colby starts with 50 bacteria in his culture and the number of bacteria doubles every 2 hours. jaquan has a different type of bacteria that doubles every 3 hours. how many bacteria should jaquan start with so that they have the same amount at the end of the day?
btw i dont want bacteria formula

I assume here that the day here is the 24 period. This means that if the bacteria doubles every 2 hours, this will happen 12 times (24/2=12).

So Colby will have 50*2*2*2*2*2*2*2*2*2*2*2*2=50*4*4*4*4*4*4=200*4*4*4*4*4=800*4*4*4*4=3200*4*4*4=204800 bacteria.

if the bacteria doubles every 3 hours, this will happen 8 times (24/3=8).

So Jaquan will have

x*2*2*2*2*2*2*2*2=x*4*4*4*4=x*16*16=x* 256

So if we want so have the same amount:

204800=x*256
x=800

So he needs to start with 800.

It is b as the correct answer is 0.088, you are able to work these out on a calculator or without a calculator if you type in the exact sum that is given for harder questions like these!