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?

What you want to do here is make 2 equations with the information provided.  first you want to define variables i will be using C= total amount of bacteria Colby has.  J=total amount of bacteria Jaquan has.    X=starting bacteria for Colby.  T1=time in 3 hour intervals.  T2=time in 2 hour intervals.  and Y=starting bacteria for Jaquan.  since we know that X=50 we won't be needing that X. 

J=Y(2)^T1 (since we are doubling the amount each time, we multiply the starting amount by 3 (every time it multiplies) to the T1 (time) power)

C=50(2)^T2

now we need them to be equal or J=C so we put each equation into one but each is on the other side of the equal sign

Y(3)^T1=50(2)^T2

since no time is given, let's use 6 hours (divisible by 3 and 2)  and we only have to figure out Y so it will fit making the equation true

Y(3)^2=50(2)^3

Y(3)^2=400

Y(9)=400
divide each side by 9 to get the Y alone

Y=4.44 repeating decimal or rounding to about 4.4.