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Population drop word problem -- help with the Algebra please

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  • #1
opus
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1. Homework Statement
You’re testing the effect of a noxious substance on bacteria. Every 10 minutes, one-tenth of the bacteria which are still alive are killed. If the population of bacteria starts with 10^6, within which period of 10 minutes will 70% of the bacteria be killed?

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The Attempt at a Solution



I tried three different ways to solve, which are all on the attached image (seemed more effective to post that so you can see my reasoning). I’d just like to know if my equation set up is incorrect or if my calculations are incorrect. Each of the three attempts is in its own box. I know I’m wrong on all three, as the solution from the book says 110-120 minutes.
 

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  • #2
andrewkirk
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The equation setup is incorrect. It looks like you are using ##x## for the number of periods. Then ##x## must be an exponent, not something that just multiplies other numbers.

To get you started, think about this: What proportion of the original bacteria is left after 20 minutes?
Now what about after 20 minutes?
Now what about 30 minutes?

With luck you will see the pattern and get the correct formula.
 
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  • #3
symbolipoint
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I guess we are not allowed to post "complete" solutions here as responses. I will give only this as guidance, so if opus want, you should work with this as best you can.

Try starting with a model like y=p*e^(-kx).
x is the count of ten-minute time periods and p is the initial count of the bacteria.
You can use some of the given information to find the value for k. The first step for that may look like this:
10^6*e^(-10k)=0.9*10^6

That's all I give here. Work with that and take it as far as you can.
 
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  • #4
opus
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Ok thanks for the responses.

Here's the reasoning that I had:
We start with a factor of 10^6, and every period of 10 minutes, that gets cut by 1/10.
So by the second period, period 2, I figured it to look like 2(10^6(1/10)), meaning that the process of cutting the 10^6 by 1/10 happened a second time (hence the 2 for period 2).

By your reasoning, in what proportion of the bacteria is left after 20 minutes, 30 minutes, etc: are we saying we are taking 1/10th of 1/10th of 10^6, or rather 10^6(1/10)^2?
 
  • #5
opus
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I guess we are not allowed to post "complete" solutions here as responses. I will give only this as guidance, so if opus want, you should work with this as best you can.

Try starting with a model like y=p*e^(-kx).
x is the count of ten-minute time periods and p is the initial count of the bacteria.
You can use some of the given information to find the value for k. The first step for that may look like this:
10^6*e^(-10k)=0.9*10^6

That's all I give here. Work with that and take it as far as you can.
What are you using k as? The period?
 
  • #6
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Here's what I think makes the most sense if I've understood correctly.

##0.7\left(10^6\right)=\left(10^6\left(\frac{1}{10}\right)\right)^x##

In words,

LHS: 70% of the starting population of 10^6 bacteria

RHS: The decay process of 10^6(1/10) happening x times.
 
  • #7
andrewkirk
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That formula won't give the correct answer because it is raising the population (##10^6## bacteria) as well as the attrition factor (##1/10##) to the power ##x##. Dimensional Analysis is helpful in avoiding such errors. We want each side of the equation to give a number of bacteria. Given the ##10^6## has units 'bacteria' (ie number of bacteria), the LHS of your equation has units 'bacteria' on the left, but units
'bacteria##{}^6## ' on the RHS. Since the units do not match, the equation cannot be correct.

With attrition problems like this, it's easier to count the number of survivors, not the number that die. The number of survivors needs to be ##10^6\times 0.30##, and that can go on the LHS of the equation. Now what do we want to raise to the power of ##x## on the RHS of the equation to represent the impact of ##x## years of population reduction? Then we need to multiply that factor on the RHS by ##10^6\ \textrm{bacteria}## to express it as a number of bacteria.
 
  • #8
symbolipoint
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opus writes the description:
You’re testing the effect of a noxious substance on bacteria. Every 10 minutes, one-tenth of the bacteria which are still alive are killed. If the population of bacteria starts with 10^6, within which period of 10 minutes will 70% of the bacteria be killed?
You could use a model based on y=p*e^(-kx) but other ways for the exponential decay model are possible.

The description you gave says, for each 10 minute period, one-tenth of the bacteria present is killed.
100%-10%=90%
or the fraction, 0.9 remains after each 10 minute period.
If x is how many ten-minute period, p the starting amount of bacteria, y the remaining amount of bacteria, then you can use a model y=p(0.9)^x.
Again see closely that I am using here, x is the count of ten-minute periods - NOT how many minutes.
Since your example gives p=10^6, you can make the more specific y=(10^6)(0.9)^x.
 
  • #9
symbolipoint
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What are you using k as? The period?
NO. k is not any period. It is a constant.
 
  • #10
OmCheeto
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Not knowing any preexisting equations, I had to solve the problem from scratch.

Solving for the first 2 ten second time periods with basic maths:
t y
0 1,000,000
10 900,000 (= 0.9 times the previous value)
20 810,000 (= 0.9 times the previous value)

I noted that the slope of y vs t changed: -10000 from t = 0 to t = 10, and -9000 from t = 10 to t = 20.
From the clues given by others, I decided to test the slope when taking the log of y

t ln(y)
0 ln(1,000,000)
10 ln(900,000)
20 ln(810,000)

This yielded equal slopes for both the periods 0 to 10 and again for 10 to 20.

From there, it was fairly easy to develop the equation that solved the problem.
It's of the form: y = mx + b, with strategically placed "ln"s of course.

It wont look like

y=p*e^(-kx)
But with some extra manipulations, it will.
 
  • #11
symbolipoint
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OmCheeto,
You appear to be linearizing the calculated results for each counted time period, but starting with the expected exponential relationship from the description would be far better. You can then do the linearization later. Another good starting point for linearizing would be if there were some actual two-D data points, from which one could draw a graph, and take logarithm of one of the coordinate values.
 
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  • #12
FactChecker
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CORRECTION: The problem asked for a reduction of 70% so the multiplier needs to be 0.3 for 30% remaining.
You can ignore the initial number of ##10^6## because 70% 30% is 70% 30% no matter how much you start with. You are looking for a multiplier that is 70% 30% (i.e. a multiplier of 0.7 0.3).

I suggest that you look at the percentage after 1 period, after 2 periods, etc. to see the pattern. Then you can make an equation for any number, n, periods or you may just want to continue by hand till you get below 70% 30%.
 
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  • #13
OmCheeto
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OmCheeto,
You appear to be linearizing the calculated results for each counted time period, but starting with the expected exponential relationship from the description would be far better. You can then do the linearization later. Another good starting point for linearizing would be if there were some actual two-D data points, from which one could draw a graph, and take logarithm of one of the coordinate values.
I went back and used your method from post #2. It worked! I didn't believe it would.
This is really simple when you know the answer.

given: y=p*e^(-kx)
find: k

Step 1: solve for k without any values. ie: : k = f(p, x, y)
Step 2: plug in numbers for time increments 10 and 20
Step 2.5: Discover you don't have y values for those time increments (should have listened to
Step 2.75: Determine y values for those time increments
Step 2.875: Perform step 2
Step 3: if the "k"s are equal, then the equation is probably valid for all values.
Step 3.5: plug in numbers for other time increments and discover that k is always the same. (up to the limits of your calculating machine)
Step 4: solve the original equation for x without values. ie: x = f(p, k, y)
Step 5: plug and chug!

ps. In my defense, had I not done the problem "my way", I'd have not discovered that "k" is the slope(m), and "p" is the y intercept(b).

exponent.maths.png
 

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  • #14
symbolipoint
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OmCheeto,
Just be careful what variables or numbers you call how. For a model as exponential using y=pe^(-kx), the number k is not a slope.

When you linearize this equation starting with taking of logarithms of both sides, then the equation as ln(y)=-kx+ln(p), is a linear equation (probably) and slope here is -k. This general example is for exponential decay. Something similar can be done for exponential growth.
 
  • #15
OmCheeto
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OmCheeto,
Just be careful what variables or numbers you call how. For a model as exponential using y=pe^(-kx), the number k is not a slope.
Well, like @opus , I was confused what "k" represented.

What are you using k as? The period?
"k" is the slope, of something.
 
  • #16
symbolipoint
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Well, like @opus , I was confused what "k" represented.



"k" is the slope, of something.
k is a constant but its use as a slope depends how the exponential equation is treated. It is not a slope until logs taken of both sides of the exponential form and the algebraic rearrangement made. Only THEN k or -k becomes the slope of a line.
 
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  • #17
Ray Vickson
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You can ignore the initial number of ##10^6## because 70% is 70% no matter how much you start with. You are looking for a multiplier that is 70% (i.e. a multiplier of 0.7).

I suggest that you look at the percentage after 1 period, after 2 periods, etc. to see the pattern. Then you can make an equation for any number, n, periods or you may just want to continue by hand till you get below 70%.
Actually, you want to get below 30% (since you want to kill off at least 70% of the initial population).
 
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  • #18
FactChecker
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Actually, you want to get below 30% (since you want to kill off at least 70% of the initial population).
Oh! I stand corrected. I will go back and correct my post.
 
  • #19
opus
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Thanks for all the responses guys. And apologies for the late response. Been busy with my classes.

So in reading all of your posts, I want to say the appropriate equation would be:

$$0.3\left(10^6\right)=10^6\left(\frac{1}{10}\right)^x$$

With this, the LHS would represent 30% of the initial population, and the RHS would represent the initial population multiplied by the "decay" factor.
This seems correct, but I don't want to say that it is because the section of the book hasn't gone over solving for variable exponents yet. Thoughts?
 
  • #20
symbolipoint
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Take another look at post #8.
If you have decay occurring, exponentially, and you KEEP 90% after each time increment, then your model would or could be done as y=p(0.9)^x. Now with this, if you want to decay from 100% to 30%, then you setup the values in the model like this:

0.3=(0.9)^x

Be reminded, x is how many increments of 10 minutes each. (Or did I completely misread, and you are keeping 10% after each time increment?)

If the model I listed here is what you need, then you could apply x at 1, 2, 3, 4, 5, etcetra and see during which increment you get your "0.3".
You can also solve the equation for x, which you will certainly be learning, probably about now...
 
  • #21
opus
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So you're saying replace (1/10) with (0.9) because the 1/10 represents what's killed off, and the 0.9 represents what's left? So we would have ##0.3\left(10^6\right)=10^6\left(0.9\right)^x##?

Not sure why this problem is giving me so much gruff. Thanks for the help.
 
  • #22
symbolipoint
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So you're saying replace (1/10) with (0.9) because the 1/10 represents what's killed off, and the 0.9 represents what's left? So we would have ##0.3\left(10^6\right)=10^6\left(0.9\right)^x##?

Not sure why this problem is giving me so much gruff. Thanks for the help.
Start with what is simple. What is the decay like? Is this KEEPING 90% after each time period? Then y=p*0.9 for the FIRST period.
What happens after two periods? y=p*0.9*0.9.
What happens after three periods? y=p*0.9*0.9*0.9.
What happens after four periods? y=p*0.9*0.9*0.9*0.9, OR using exponents, x counting each time period, y=p(0.9)^4.
Variable number of time periods? y=p*(0.9)^x.
 
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  • #23
opus
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Ok I definitely understand that reasoning. But why are we using (0.9) rather than (1/10)? Is it because when we use (0.9), we're talking about whats left, and when we use (1/10), we're talking about what was killed? And of course we want to know how many bacteria are left.

So our equation should look like ##0.3\left(10^6\right)=10^6\left(0.9\right)^x##, with x being the number of ten minute periods?
 
  • #24
symbolipoint
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Ok I definitely understand that reasoning. But why are we using (0.9) rather than (1/10)? Is it because when we use (0.9), we're talking about whats left, and when we use (1/10), we're talking about what was killed? And of course we want to know how many bacteria are left.

So our equation should look like ##0.3\left(10^6\right)=10^6\left(0.9\right)^x##, with x being the number of ten minute periods?
Look literally what the problem description is. LOOK and THINK as literally as possible. What is said in the description? Is the bacteria losing 10% every ten minutes? Is the bacteria losing 90% every ten minutes? This becomes a matter of Reading Comprehension.
 
  • #25
symbolipoint
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Ok I definitely understand that reasoning. But why are we using (0.9) rather than (1/10)? Is it because when we use (0.9), we're talking about whats left, and when we use (1/10), we're talking about what was killed? And of course we want to know how many bacteria are left.

So our equation should look like ##0.3\left(10^6\right)=10^6\left(0.9\right)^x##, with x being the number of ten minute periods?
Here is what you wrote as the description in post #1:

Every 10 minutes, one-tenth of the bacteria which are still alive are killed. If the population of bacteria starts with 10^6, within which period of 10 minutes will 70% of the bacteria be killed?
Now, if x is the counting of the ten-minute increments, y is living bacteria after x time, p is amount of bacteria at time x=0, you should be able to read ( and maybe reread) the description and form the exponential relationship and choose the correct base for the x.

One-tenth of the bacteria is killed after each ten-minute time increment. So what fraction then still lives?
 

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