1
$\begingroup$

"A certain type of bacteria doubles every 6.5 hours. If there were 60 bacteria to start with, what is the hourly growth rate of the bacteria? How many bacteria will there be after a day and a half? Solve using algebra."

Another person was asking about this on here but I couldn't find the answer. I've come up with the equation $\ t = b2^{x/6.5}$ but I'm not sure if this is right/where to go from here.

$\endgroup$
6
  • 3
    $\begingroup$ Your exponent is correct but $t$ is usually used as the time variable, not $x$. Your equation should be of the form $b(t)=b(0)\cdot2^{t/6.5}$ where $t$ is time in hours and $b(t)$ is the number of bacteria at time $t$. $\endgroup$ Commented Apr 19, 2019 at 15:27
  • $\begingroup$ Ah. Do I just begin to substitute in the provided numbers from there? $\endgroup$
    – andrewt
    Commented Apr 19, 2019 at 15:27
  • $\begingroup$ Yes, that would be the approach. $\endgroup$ Commented Apr 19, 2019 at 15:28
  • $\begingroup$ You can also use that $$N(t)=N_0e^{\lambda t}$$ $\endgroup$ Commented Apr 19, 2019 at 15:28
  • 1
    $\begingroup$ The hourly rate is $b(1)/b(0)$ $\endgroup$
    – Andrei
    Commented Apr 19, 2019 at 15:28

1 Answer 1

0
$\begingroup$

Hint: In general the equation is $y=y_0\cdot a^t$, where $y$ is the poputlation of the bacteria after $t$ hours and $y_0=60$ is the initial population. Now you can deduce an equation from the following information:"A certain type of bacteria doubles every $6.5$ hours"

$$120=60\cdot a^{6.5}$$

Solve the equation for $a$ and obtain the function of the population.

$\endgroup$
5
  • $\begingroup$ why is it "$\ 120 =$"? where did you get that? $\endgroup$
    – andrewt
    Commented Apr 19, 2019 at 15:34
  • $\begingroup$ 120 is double of 60. $\endgroup$ Commented Apr 19, 2019 at 15:37
  • $\begingroup$ @andrewt Please give a reply if it is all clear now. $\endgroup$ Commented Apr 19, 2019 at 16:02
  • $\begingroup$ All clear. Thank you for the answer, sorry for the long response time. $\endgroup$
    – andrewt
    Commented Apr 20, 2019 at 3:53
  • $\begingroup$ No, problem. The reply wasn´t late at all. But a late reply is 100 times better than no reply. $\endgroup$ Commented Apr 20, 2019 at 13:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .