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Mathematics 15 Online
OpenStudy (anonymous):

a certain bacteria will triple in 6 hours. if the final count is 8 times the original cont , how much time has passed?

OpenStudy (anonymous):

*count

OpenStudy (anonymous):

let the first count be a and the next count ar a,ar,ar^2,ar^3.... \[a_n=ar^{n-1}\] where n is a period after 6 hours

OpenStudy (anonymous):

okay .

OpenStudy (anonymous):

\[a_n=a(3)^{n-1}\]

OpenStudy (anonymous):

why is r =3 ?

OpenStudy (anonymous):

because every time we triple (3) the count,we suppose this occurs every 6 hours

OpenStudy (anonymous):

okay.

OpenStudy (anonymous):

\[8a=a3^{n-1} \implies 8=3^{n-1} \implies n-1=\log_38\]

OpenStudy (anonymous):

so log sub 3 8 isnt my answer ., its what i plug into the calculator ?

OpenStudy (anonymous):

yes so how much is n

OpenStudy (anonymous):

1.8927892607 ?

OpenStudy (anonymous):

+1

OpenStudy (anonymous):

in an hour ?

OpenStudy (anonymous):

n-1=1,892789607 n=2,89... thiS mean we have to multily by 6 17,35h

OpenStudy (anonymous):

@RadEn

OpenStudy (anonymous):

and so , what would be the final step?

OpenStudy (anonymous):

17,35 hours=17 hours 21 minutes

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