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a certain bacteria will triple in 6 hours. if the final count is 8 times the original cont , how much time has passed?
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*count
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
okay .
\[a_n=a(3)^{n-1}\]
why is r =3 ?
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because every time we triple (3) the count,we suppose this occurs every 6 hours
okay.
\[8a=a3^{n-1} \implies 8=3^{n-1} \implies n-1=\log_38\]
so log sub 3 8 isnt my answer ., its what i plug into the calculator ?
yes so how much is n
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1.8927892607 ?
+1
in an hour ?
n-1=1,892789607 n=2,89... thiS mean we have to multily by 6 17,35h
@RadEn
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and so , what would be the final step?
17,35 hours=17 hours 21 minutes
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