Question

I need to solve this equation for t:
0.002=0.1+e^(t/2000), which after subtracting 0.1 from both sides I am left with
-0.098=e^(t/2000). I don't have a clue how to solve it. Do I need to take the Ln of both sides?

Answer 1

yes

Answer 2

ah wait you'll get a negative argument for the ln

Answer 3

It does appear like you'll get some complex logarithm

Answer 4

maybe I have something wrong, this is the second part of a problem, which was right for the first question.

Answer 5

Did you every use complex numbers to solve these? Or maybe you did do a wrong calculation if you expect a real (non-complex) answer

Answer 6

ever*

Answer 7

No I don't think I should end up with complex numbers, I'm looking for an exact time in minutes.

Answer 8

The problem goes like this: A swimming pool whose volume is 10,000 gal contains water that is 0.001% chlorine. Starting at t=0, city water containing 0.001% chlorine is pumped into the pool at a rate of 5 gal/min. The pool water flows out at the same rate. What is the percentage of chlorine in the pool after 1 hour? When will the pool water be 0.002% chlorine?

Answer 9

is the % of chlorine initially and being pumped the same?

Answer 10

yes, the input and output are the same rate.
I was able to solve the percentage of the chlorine in the pool after 1 hour was I got the general solution x(t)=0.1+0.9/e^(t/2000) and then using that to solve for x(60min). which I got 0.0097% which was correct,

Answer 11

SOrry I meant the percentage %, not the rate of flow

Answer 12

yes the pool contains 0.01% chlorine initially. Is that what your asking?

Answer 13

ah ok.

Answer 14

What's the 2nd part of the problem then

Answer 15

so the second part of the question that I am having trouble with is, When will the pool water be 0.002% chlorine? which I just plugged 0.002% in for x(t) and got 0.002=0.1+0.9/e^(t/2000). But now I'm wondering if that is right

Answer 16

Hm I get:
\(\Large 0.1+0.9e^{-t/2000}=0.973 gallon\)
but then
\[ \Large \frac{0.973}{10,000}=0.0000973=>0.00973\%\]
So similarly, for 0.002%, it should give 0.2 gallon to plug into the equation.

Answer 17

interesting, I just lucked out because I just divided 0.973 by 100. So my solution is wrong.
But why, or how did you multiply e^(t/2000) to C?

Answer 18

Not sure what you mean? Did you get a step like: \[\ln|0.1-x|=\frac{-t}{2000}+C\]

Answer 19

I was referring to the general solution, When I solved for (mu) and then used mu to get the general equation. \[d/dt[(x)e^(t/2000)]=0.00005e^(t/2000)\]

Answer 20

because I integrated both sides which gave me xe(t/2000)=0.1e^(t/2000)+C, so I divided both sides by e^(t/2000) which is how I got x=0.1+C/e^(t/2000).

Answer 21

Hm, I did the following way:
\[ \frac{dx}{dt}=0.00005-5\frac{x}{10000}\\
=0.00005-0.0005x\\
=0.0005(0.1-x)\\
\frac{1}{0.1-x}\frac{dx}{dt}=0.0005\\
\frac{1}{0.1-x}dx=\frac{1}{2000}\,dt\\
-\ln{|0.1-x|}=\frac{t}{2000}\\
\ln{|0.1-x|}=\frac{-t}{2000}+C\\
|0.1-x|=e^{-t/2000}e^C\\
0.1-x=\pm e^Ce^{-t/2000}=Ae^{-t/2000}\\
x=0.1-Ae^{-t/2000}\]
Now at the intial condition, \(x(0)=1\), since 0.01% into gallon, just like before, is 1
so: \(x(0)=1=0.1-Ae^0\implies A=-0.9\)
, soo: \[ x=0.1+0.9e^{-t/2000}\]

Answer 22

wow, that is much easier than what I was trying to do. I was trying to set it up the way my professor showed us by using the standard form, getting p(t), and Q9t), and mu(t). Thank you so much.

Answer 23

:) you're welcome.

Answer 24

You shouldn't get a negative argument for the ln now :) especially since you have to be careful about the percentage thing .

Answer 25

right, thanks again.