Law of uninhibited growth

[alyren]

A bacterial culture has an initial population of 10,000. if its population declines to 3000 in 2 hours, when will its population be 900? assume that the population decreases according to the exponential model.

is this equation set up right?
10000=3000e^k2
i solved for k

then 900=e^ln(10000/3000)/2*t
then solve for t?

 

[Mrspi]

A bacterial culture has an initial population of 10,000. if its population declines to 3000 in 2 hours, when will its population be 900? assume that the population decreases according to the exponential model.

is this equation set up right?
10000=3000e^k2
i solved for k

then 900=e^ln(10000/3000)/2*t
then solve for t?

I think you may have the 3000 and the 10000 in the wrong places. Isn't the INITIAL quantity 10000?

Final amount = Initial amount * e^(k t)

You need to find "k" first....

 

[alyren]

so it will be 3000=10000e^k2
then find k

then 900=e^Ln(watever k is)*t?
then solve for t?

 

[mmm4444bot]

3000 = 10000 e^k2

Your typing means this:

\(\displaystyle 3000 \;=\; 10000 e^{k} \cdot 2\)

because you did not type grouping symbols around the exponent.

What you mean is this:

3000 = 10000 e^(2k)




then 900 = e^Ln(watever k is)*t No

Do not take the natural log of k.

Where is the initial amount ?

Follow the model that Mrs. Pi provided.

 

[alyren]

ok let me make this more clear

the equation
3000 = 10000 e^(2k)

divide 10000 both side

(3000/10000)=e^(2k)

then i Ln both side to get rid of e

Ln(3000/10000)=2k

divided by 2 both side

k=(ln3000/10000)/2

for 2nd equation
when will its population be 900?

900=e^(k*t)
900=e^[(ln3000/100000)/(2) *t]
then solve for t

is this right?
i hope this make it more clear.

 

[mmm4444bot]

3000 = 10000 e^(2k)

(3000/10000)=e^(2k) ? At this point, you could reduce the Rational number 3000/10000.

3/10 = e^(2k)

ln(3000/10000) = 2k

k = (ln3000/10000)/2 ? These grouping symbols are not correct.

The input to ln(x) is 3000/10000, not 3000.

k = ln(3000/10000)/2



when will its population be 900?

900=e^(k*t) ? You are still not using the proper model.

Where is the initial amount ?

900 = 10000 e^(kt)

Did they ask you to do any rounding ?

There's nothing wrong with using the exact value of k, but in real-world applications we usually write a decimal approximation.

For example, we can write:

k = -0.60199 (I rounded k to five decimal places)

Using this approximate value for k yields the following.

10000 e^([-0.60199][2]) = 2999.9784

which seems close enough to 3000, to me.

Otherwise, if you stick with exact values, your answer for t will look like this:

t = 2[ln(9/100)/ln(3/10)]

Again, if that's what they want, then that's okay.

Otherwise, it might be easier for you to use decimal approximations.

Cheers ~ Mark

 

[alyren]

no the question did not ask for decimal rounding.

i just want to see if my equation is set up right.

 

[mmm4444bot]

i just want to see if my equation is set up right.

I understand.

No, your equation is not set up correctly.

It should be:

900 = 10000 e^(kt)

and replace the symbol k with either a decimal approximation or the exact expression.

 

[alyren]

oh ok, i though i only need the initial for the first equation to find out what the k is then plug in k for the 2nd equation to find out what t is. good to know, thanks

ok so after i solve the equation

3000=10000e^(2k)

900=10000e^(kt)

i got t=4 hours, is this right?

maybe it make sense to some people, but it doesn't make sense to me that population 10000 to 3000 in 2 hours and 10000 to 900 4 hours..