colby and Jaquan are growing bacteria in an experiment in a laboratory. Colby starts with 50 bacteria in his culture and the num
ber 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.
divide the original price by 2 since it is half off. that gives you 6.44. so you would multiply 6.44 by 5, since allison wants 5 cases. that gives you 32.20 dollars