Question

A bacteria population experiences continuous growth. If there were 2,000 bacteria present at the start of the experiment, how many did it take for there to be 6,000 bacteria if the growth rate was 7% per hour? I will grant you a fan and a medal.

Answer 1

hI!!

Answer 2

start with the equation for growth and 7% per hour, do you know it?

Answer 3

Hi!

Answer 4

2,000 * 0.07?

Answer 5

not quite

Answer 6

to increase a number by \(7\%\) multiply it by \[100\%+7\%=107\%=1.07\]

Answer 7

to do it say \(t\) times, multiply it by \(1.07^t\)

Answer 8

OH! So were timesing it by 1.07, okay.

Answer 9

right and your job is to solve \[2000\times (1.07)^t=6000\]any clues ?

Answer 10

"no clue" is ok i can show you, just asking

Answer 11

Yep, totally clueless.

Answer 12

ok first step is obvious probably , divide by 2000

Answer 13

\[1.07^t=3\]

Answer 14

so far so good?

Answer 15

did i lose you there?

Answer 16

2.80?

Answer 17

Is that what t is?

Answer 18

i doubt it triples that fast

Answer 19

you need either a fancy calculator or logs so solve this
does your class do logs and exponents?

Answer 20

Yes.

Answer 21

are you familiar with the "change of base" formula"?

Answer 22

No...

Answer 23

\[b^x=A\iff x=\frac{\ln(A)}{\ln(b)}\] the log of the total divided by the log of the base

Answer 24

so \[1.07^t=3\iff t=\frac{\ln(3)}{\ln(1.07)}\] and a calculator

Answer 25

what class has conic sections and logs? wowee

Answer 26

Pre-Calc.

Answer 27

really mixing it up
did you get an answer ?

Answer 28

No.

Answer 29

16.24