Question 1180947
.
A certain type of bacteria, given a favorable growth medium, doubles in population every 6.5 hours. 
Given that there were approximately 100 bacteria to start with, how many bacteria will there be 
after one and a half days?
~~~~~~~~~~~~~~~~


<pre>
With the doubling peroiod of 6.5 hours and initial population of 100 bacteria,


the current population, as the function of time, is  n(t) = 100*2^(t/6.5),  where t is the time in hours.



To find the population in 1.5 days = 1.5*24 = 36 hours,  the fotmula is


    n(36 hours) = 100*2^(36/6.5).


Use your calculator.
</pre>

------------


Solved.


Is everything clear to you in my explanation ?


If you still have questions, do not hesitate ask me.



/////////////



To see many other similar and different solved problems on bacteria growth, &nbsp;look into the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/logarithm/Bacteria-growth-problems.lesson>Bacteria growth problems</A> 

in this site.


Also, &nbsp;you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lesson is the part of this online textbook under the topic "<U>Logarithms</U>".



Save the link to this online textbook together with its description


Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson


to your archive and use it when it is needed.