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There were 3000 bacteria in a petri dish 10 hours ago. Now there are only 2000. How...

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oyechalphut | Student, Grade 10 | Honors

Posted March 30, 2013 at 11:01 PM via web

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There were 3000 bacteria in a petri dish 10 hours ago. Now there are only 2000.

How would you calculate how many bacteria there will be 5 hours later and at what percentage the bacteria is decaying per hour. 

Tagged with functions, long question, math

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mariloucortez | High School Teacher | (Level 3) Adjunct Educator

Posted March 31, 2013 at 12:07 AM (Answer #1)

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To solve this, you can use the formula for exponential decal/growth:

`F = Pe^(rt)`

 where F = value at time, t

           P = initial value (value at time, t=0)

           r = growth/decay factor (+ if it is a growth factor, - if it is a decay)

           t = time, in hours (in this case)

First, identify the given.

P = 3000 @ t = 0 (10 hours ago)

F = 2000 @ t = 10 (t = now, after 10 hours)

t = 10

Plug-in in the formula:

`F = Pe^(rt)`

`2000 = 3000e^(r*10)`

Divide both sides by 3000, so e^(rt) is isolated, since it contains an unknown, the r (decay factor). It is a decay factor because, the initial value is 3000 then it dropped to 2000.

`2000/3000 = (3000e^(10r))/3000`

 `2/3=e^(10r)`

Take the natural logarithm of both sides and use the properties:

`ln e = 1` and;

`ln(x^n) = nln(x)`

So,

`ln(2/3) = ln(e^(10r))`

`ln(2/3) = 10rln(e)`

`-0.4055 = 10r`

Divide both sides by 10 to solve for r. (I just rounded ln(2/3) to 4 decimal places, it is better to get the exact result by saving it in your calculator for accuracy).

`-0.4055/10 = (10r)/10`

`r =-0.04055`

No that you have the decay factor, calculate the bacteria after 5 hours using the same formula.

P = 2000 (This is our initial value because we move our time to 'now'. You can still use 3000 and get same value but your time, t would be equal to 15 instead.)      

F = ?

t = 5

`F = 2000e^(-0.04055*5)`

`F = 1632.99316`

Therefore there are 1632.99316 bacteria after 15 hours.

As per how the percentage of bacteria is decaying per hour, you can get the bacteria at t = 1 using 2000 as the initial value.

`F = 2000e^(-0.04055*1)`

`F =1920.529002`

Then to get the percentage:

`% = (2000-1920.529002)/2000*100`

` `

`% = 3.9735%`

Therefore almost 4% of bacteria decayed per hour.      

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