# doubling bacteria

• January 26th 2010, 07:22 PM
ryan18
doubling bacteria
A bacteria culture starts with 800 bacteria and doubles in size every half hour.

I have already found the formula to be:
$f(x)=800(2)^{2t}$

I just am having trouble doing this question...

(d) Graph the population function and estimate the time for the population to reach 70,000. (Give your answer to the nearest hundredth.)
• January 26th 2010, 07:30 PM
pickslides
Quote:

Originally Posted by ryan18
A bacteria culture starts with 800 bacteria and doubles in size every half hour.

I have already found the formula to be:
$f(x)=800(2)^{2t}$

I just am having trouble doing this question...

(d) Graph the population function and estimate the time for the population to reach 70,000. (Give your answer to the nearest hundredth.)

Or should it be $f(x)=800(2)^{t}$ ?
• January 26th 2010, 07:32 PM
Jhevon
Quote:

Originally Posted by ryan18
A bacteria culture starts with 800 bacteria and doubles in size every half hour.

I have already found the formula to be:
$f(x)=800(2)^{2t}$

I just am having trouble doing this question...

(d) Graph the population function and estimate the time for the population to reach 70,000. (Give your answer to the nearest hundredth.)

this is just an $a^x$-graph, where a is a constant. they have a general form, they look like the $e^x$-graph which you should know. the difference here is the y-intercept is 800.

for the estimate, plug in 70,000 for f(x) and solve for t. round to the nearest hundredth
• January 26th 2010, 07:43 PM
ryan18
I dont know what to do once I get it down to

$87.5 = 2^{2t}$

Do I take the Log_2 of each side to get 2t? How do I compute that for the 87.5?
• January 26th 2010, 07:51 PM
Stroodle
$\frac{175}{2}=2^{2t}$

$log_2\left ( \frac{175}{2}\right )=2t$

$t=\frac{1}{2}log_2\left ( \frac{175}{2}\right )$

$t\approx 3.23$
• January 26th 2010, 07:52 PM
Jhevon
Quote:

Originally Posted by ryan18
I dont know what to do once I get it down to

$87.5 = 2^{2t}$

Do I take the Log_2 of each side to get 2t? How do I compute that for the 87.5?

you can use the change of base formula: remember, $\log_a b = \frac {\log_c b}{\log_c a}$

so, $\log_2 x = \frac {\log x}{\log 2}$, and you can find $\log$ on your calculator.
• January 26th 2010, 07:53 PM
pickslides
$87.5 = 2^{2t}$

$\log_287.5 = 2t$

$t = \frac{\log_287.5}{2}$

Use calculator on RHS.