"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.

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    $\begingroup$ 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 $t$. $\endgroup$ – John Wayland Bales Apr 19 at 15:27
  • $\begingroup$ Ah. Do I just begin to substitute in the provided numbers from there? $\endgroup$ – andrewt Apr 19 at 15:27
  • $\begingroup$ Yes, that would be the approach. $\endgroup$ – John Wayland Bales Apr 19 at 15:28
  • $\begingroup$ You can also use that $$N(t)=N_0e^{\lambda t}$$ $\endgroup$ – Dr. Sonnhard Graubner Apr 19 at 15:28
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    $\begingroup$ The hourly rate is $b(1)/b(0)$ $\endgroup$ – Andrei Apr 19 at 15:28

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.

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

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