A bacteria culture that doubles each hour has an initial population of 7 cells. Write an hourly equation modeling this information. How many cells are present 24 hours later?

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Answer 1 · 2018-01-22T16:02:36

#117440512# cells

We need to find an exponential equation of the form:

#A(t)=ae^(kt)#

Where:

#a# is the initial amount.

#k# is the growth/decay factor.

#t# is the time. In this case hours.

#A(t)# is the amount after time #t#

From information given:

We have #a=7#

We are told this doubles each hour, so after #1# hour we would expect there to be 14 cells.

#A(t)=14#

#t=1#

Using these values in the equation.

#14=7e^(k(1))#

We now solve this for #k#:

Divide both sides by 7:

#2=e^(k(1))#

Taking natural logarithms of both sides:

#ln(2)=kln(e)color(white)(8888)# ( #ln(e) =1#, the log of the base is always 1)

#ln(2)=k#

Now our equation is:

#A(t)=7e^((ln(2))t)color(white)(88888)# ( #e^ln(2) = 2# )

#A(t)=7(2)^t#

After 24 hours

#t=24#

#:.#

#A(t)=7(2)^24=117440512# cells