Question

Suppose that the concentration of a bacteria sample is 30,000 bacteria per milliliter. If the concentration triples in 4 days, how long will it take for the concentration to reach 51,000 bacteria per milliliter? (In days)

Answer 1

It's the 6th time I get the problem wrong. Each time I get a similar problem, I don't get it.

Answer 2

@whpalmer4
@thomaster

Answer 3

@e.mccormick
@iambatman
@jigglypuff314

Halp

Answer 4

@AccessDenied
@Ashleyisakitty
@mathmale

Answer 5

@surjithayer @zepdrix @satellite73 @mathslover if you're not too busy could your guys try to help please? :3

Answer 6

I'm trying to work it out, although it's a bit confusing. Is there any answer choices?

Answer 7

No there aren't. I know it's a Ca^x problem, just don't get what I'm doing wrong

Answer 8

Can you show what you've done so far?

Answer 9

Well I tried setting it up. C is 30,000

Answer 10

\[C(30,000)^4 = 120,000\]I think we can work around this way.

Answer 11

Oh whoops, that 120,000 is supposed to be 90,000

Answer 12

How did you set that up, m8?

Answer 13

\(C\times a^x = \rm new~thing\)

And \(a= 30,000\), \(x = 4\). It's given that the thing triples so the final quantity is \(90,000\)

Answer 14

So we can solve for \(C\) from that and plug into\[C\times (30,000)^x = 51,000\]

Answer 15

u alrite m8?

Answer 16

Wouldn't C be 30,000? As the original thing?
I thought A was the growth factor

Answer 17

Oooh.

Answer 18

Luce do you get unlimited guesses?
or does it give you a new question after you've attempted this one?

I have an answer but I'm not sure if I did it correctly :o

Answer 19

OK, then use the equation\[30,000 \times a^4=90,000\]

Answer 20

I get three or two tries I believe. After I get it wrong, I get a similar question. So unlimited tries at a type of problem, a couple at the specific problem.

Answer 21

OK, I promise this one was right.

Answer 22

\[a ^{4}=3\]

Answer 23

Right... log em up.

Answer 24

Natural log, amirite?

Answer 25

any base would work.

Answer 26

So I was attempting this,\[\Large\rm B(t)=30000(a)^{kt}\]Bacteria as a function of time.
Triples every 4 days, so a=3, and then we just adjust the days.\[\Large\rm B(t)=30000(3)^{t/4}\]So that should give triple every 4 days. :o
This is kinda the same thing Parth was doing I guess.
His approach is more thorough though.

And then just let \(\Large\rm B(t)=51000\) and solve for t.

Answer 27

So wait, what I'm I setting this up as Parth? Kinda confused.
Also trying Zep

Answer 28

\[a^4 = 3\]\[\Rightarrow \log(a^4) = \log(3)\]\[\Rightarrow 4 \log(a) = \log(3)\]\[\Rightarrow \log(a) = \log(3)/4\]\[\Rightarrow a = 10^{\log(3)/4} = 10^{0.119}= 1.31\]All credits go to my calculator.

Answer 29

Now that we have found \(a\), we just plug it into the original equation\[30,000\times (1.31)^x = 51,000\]

Answer 30

Alright so, 1.31^x=1.7?

Answer 31

Right.

Answer 32

Log'ing them now

Answer 33

(x)0.2700=0.5306

Answer 34

So far so good?

Answer 35

I'd think so, all right.

Answer 36

x=1.9651

Answer 37

All righty.

So it'd take almost 2 days for 51,000.

Answer 38

I did something wrong in the math. The answer was 1.93

Answer 39

Hmm, seems like we lost significant digits in the process. Let's not calculate the value of any logarithm and take it down to the end.

With that, we have the following:\[a^4 = 3\]\[\log (a) = \log(3)/4\]\[a = 10^{\log(3)/4}\]So plugging this thing into the original equation...\[\left(10^{\log(3)/4}\right)^x = 1.7\]Messy expression, yes, but we'll have to calculate with the exact value when dealing with exponents.

I should have realized this earlier.

Answer 40

Have you lost all of your attempts? If so, I'd like to apologize.

Answer 41

Yes, don't worry it's okay. You want help me through the similar problem? My got it's the seventh time

Answer 42

I hope I can. :)

Answer 43

Suppose that the concentration of a bacteria sample is 20,000 bacteria per milliliter. If the concentration triples in 4 days, how long will it take for the concentration to reach 32,000 bacteria per milliliter? (In days)

Answer 44

Is a^4=3 again?

Answer 45

Oh, this question is very similar.

Yes - that's right.

Answer 46

I'm still confused as to where we went wrong in the math. The problem says to round all intermediate numbers to the nearest thousandth, is that it?

Answer 47

Ah, that is the problem. We didn't know how to round-off. OK, we'll now do it to the nearest thousandth.

Answer 48

So, 4log(a)=log(3)?

Answer 49

Then log(a)=(log(3))/4

Answer 50

Yes.

Answer 51

log(a)=0.119

Answer 52

Exactly. (OK, not exactly - just rounded off :P)

Answer 53

So how do I take the log out of log(a)?

Answer 54

\[\rm \log(a) = something \Rightarrow a = 10^{\large something}\]Definition of log.

Answer 55

That's how we did it in the previous question.

Answer 56

so 10^0.119=1.315

Answer 57

Seems good.

Answer 58

a=1.315?

Answer 59

Yes.

Answer 60

Then 20,000(1.315^x)=32,000

Answer 61

1.315^x=1.6

Answer 62

Right.

Answer 63

(x)log(1.315)=log(1.6)

Answer 64

Mhm, mhm.

Answer 65

x=1.716, final answer being 1.71?

Answer 66

I guess you should round-off the final answer to the nearest thousandth as well.

Answer 67

Omg guess what

Answer 68

I'm sorry I got another one wrong. :(

Answer 69

No silly, you da man

Answer 70

lulz k

Answer 71

I think we did something slightly wrong in the math again though. I put 1.71 as the answer, though when you rounded off it should have been 1.72. lmao
Anyways 1.71 was right

Answer 72

Alrighty then!