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.
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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`
Take the natural logarithm of both sides and use the properties:
`ln e = 1` and;
`ln(x^n) = nln(x)`
`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`
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)`
Then to get the percentage:
`% = (2000-1920.529002)/2000*100`
`% = 3.9735%`
Therefore almost 4% of bacteria decayed per hour.
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