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. let x equal number of days. what two equations will represent colby's growth and jaquan's?

I was stuck on this problem too but I finally figured it out. One major hint was let x= the number of DAYS. So taking both of the growth rates into account the solution is this.

Colby:
He starts with 50 bacteria so that is your a value or y intercept. Keeping in mind x is the number of days you divide the number of hours in a day (24) by the time it takes to double (2 hours) and get 12. So the equation for Colby is y=50*2^12x

Jaquan:

You do the same thing except with Jaquan's part of the equation and divide 24 by 3 to get 8. The solution for Jaquan is y=50*2^8x

Colby:He starts with 50 bacteria so that is your a value or y intercept. Keeping in mind x is the number of days you divide the number of hours in a day (24) by the time it takes to double (2 hours) and get 12. So the equation for Colby is y=50*2^12x
Jaquan:
You do the same thing except with Jaquan's part of the equation and divide 24 by 3 to get 8. The solution for Jaquan is y=50*2^8x

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