Question

The number of bacteria present in a culture after t minutes is given as b=100e^kt There are 6639 bacteria present after 5 minutes. Find k. Explain how you solve this problem.

Answer 1

i guess you are supposed to solve
\[\large 6639=100e^{5k}\]for \(k\)

Answer 2

the explanation is that you know what \(b\) is when \(t=5\)

Answer 3

do you know how to solve it? it take only 3 steps

Answer 4

no :(

Answer 5

ok lets to it step by step in case you have to do it on your own some time

Answer 6

step 1 is divide by 100

Answer 7

you get
\[\large 66.39=e^{5t}\]
good so far?

Answer 8

yes

Answer 9

so log66.39=5t?

Answer 10

yes

Answer 11

t=.37?

Answer 12

no k is .37

Answer 13

i didn't check
it is
\[\frac{\ln(66.39)}{5}\] let me check it

Answer 14

that is not what i get
i take it you are using a calculator

Answer 15

yes

Answer 16

what kind?
if you don't have to do this on a test i can show you here, but if you do we should make sure you can do it correctly on your calculator

Answer 17

just google calc

Answer 18

http://www.wolframalpha.com/input/?i=ln%2866.39%29%2F5

Answer 19

i get .36 actually

Answer 20

...weird

Answer 21

send me the link

Answer 22

https://www.google.com/search?q=calc&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a&channel=fflb

Answer 23

https://www.google.com/#q=ln%2866.39%29%2F5

Answer 24

https://www.google.com/search?q=calc&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a&channel=fflb

Answer 25

ok

Answer 26

i am going to guess you did not end the parentheses
just a guess

Answer 27

idk...what happened there

Answer 28

yeah

Answer 29

thought so

Answer 30

can you help me with another?

Answer 31

sure

Answer 32

The time required to grow a certain bacteria in a culture beginning with 100 bacteria is lnb-ln100/1.532, where B is the number of bacteria and t is the time in hours.

How much time is required to grow a culture of 15,600 bacteria? Show the steps you used to find the number of hours.

Answer 33

there is no t in your expression

Answer 34

t=lnb-ln100/1.532

Answer 35

oh ok so i guess we have to solve
\[t=\frac{\ln(1500)-\ln(100)}{1.523}\] if i am reading your expression correctly
that is just a straight up calculation
nothing really so solve for

Answer 36

it is the whole thing over \(1.523\) right?

Answer 37

yes

Answer 38

http://www.wolframalpha.com/input/?i=%28ln%281500%29-ln%28100%29%29%2F%281.523

Answer 39

look reasonable?

Answer 40

yes

Answer 41

k that was even easier, a direct substitution

Answer 42

thank you sooo much

Answer 43

youve been very helpful

Answer 44

your quite welcome