# Thread: Natural logarithm problem involving e, solve for k

1. ## Natural logarithm problem involving e, solve for k

25 = 100 / (1 + 9e^(-2k))

First I multiplied both sides by (1 + 9e^(-2k)) and got

25 + 225e^(-2k) = 100 + 900e^(-2k)

Then I subtracted 225e^(-2k) from 900e^(-2k) and 100 from 25 and got

-75 = 675e^(-2k)

I divided -75 by 675 and got

-1/9 = e^(-2k)

Now I think I'm supposed to take the natural log (ln) of both sides, but when I take the ln of -1/9 my calculator says "error: nonreal answer" What am I doing wrong?

2. Originally Posted by WahooMan
25 = 100 / (1 + 9e^(-2k))

First I multiplied both sides by (1 + 9e^(-2k)) and got

25 + 225e^(-2k) = 100 + 900e^(-2k)

Then I subtracted 225e^(-2k) from 900e^(-2k) and 100 from 25 and got

-75 = 675e^(-2k)

I divided -75 by 675 and got

-1/9 = e^(-2k)

Now I think I'm supposed to take the natural log (ln) of both sides, but when I take the ln of -1/9 my calculator says "error: nonreal answer" What am I doing wrong?
Math errors my friend. You went off the wire at the start! You multiplied way incorrectly.

$25 = 100/(1+9e^{-2K})$

$25(1+9e^{-2K}) = 100$

$1+9e^{-2K} = 4$

$e^{-2K} = 3/9 = 1/3$

Solve from there

3. Got it. k=0.55 Thank you so much.

4. When I plug k=0.55 into 100 / (1 + 9e^(-2k)) should the graph be a horizontal line at y=25.026? If so, why?

Edit: Nevermind.. stupid mistake.

5. Originally Posted by WahooMan
When I plug k=0.55 into 100 / (1 + 9e^(-2k)) should the graph be a horizontal line at y=25.026? If so, why?
I won't go into the specifics of this question, but I think you can look at what you have and discern what it should look like.

When we graph, we graph the function. For example,

$y=x^2$

Is the parabola centered at the origin going upwards. If we were to graph this then at specific points we have both an x and a y value (on a graph of the x and y domain).

In the case of your question if we graph in the X and Y domain, all we have is a constant value for Y that doesn't change with respect to X. In fact, it is completely independent of X! Therefore, we should have a straight line at that point.

HOWEVER, if you are graphing Y vs K, then this is not the case. And again we would have a particular point of K/Y but not a line.

6. I need to graph this function:

q(t) = 100 / (1 + 9e^-kt) where k = 0.55

The graph I am getting is a curve concave upwards with a y intercept at y = 100 and a horizontal asymptote at y = 0. Is this right?

7. Originally Posted by WahooMan
I need to graph this function:

q(t) = 100 / (1 + 9e^-kt) where k = 0.55

The graph I am getting is a curve concave upwards with a y intercept at y = 100 and a horizontal asymptote at y = 0. Is this right?
See that's kind of important. I didn't know you already subbed in for time! lol

Um...not entirely

Look at your denominator. Where is this min? Well it is min when t=0, why? because the part

$1 + 9e^-kt = 1 + 9/(e^{kt})$

Which you can see only gets smaller as time increases. And this is the denominator of our fraction. So smaller denominator --> bigger number

So lets take this to the limit. What happens to our function as t approaches infinity? Well our equation goes to 100. So we have an aysmtote at 100, not at 0.

When t=0 the interscept is 10, not 100.

8. Originally Posted by AllanCuz
See that's kind of important. I didn't know you already subbed in for time! lol

Um...not entirely

Look at your denominator. Where is this min? Well it is min when t=0, why? because the part

$1 + 9e^-kt = 1 + 9/(e^{kt})$

Which you can see only gets smaller as time increases. And this is the denominator of our fraction. So smaller denominator --> bigger number

So lets take this to the limit. What happens to our function as t approaches infinity? Well our equation goes to 100. So we have an aysmtote at 100, not at 0.

When t=0 the interscept is 10, not 100.
Hmm.. This is what I have in my calculator at the y= screen:

y1 = (100)/(1+9e^(-0.55)(X))

What am I putting in wrong?

9. Originally Posted by WahooMan
Hmm.. This is what I have in my calculator at the y= screen:

y1 = (100)/(1+9e^(-0.55)(X))

What am I putting in wrong?
Are you making sure you put in that negative? It is crucial because that take the e and puts it underneath the 9. Also, your graphing calculator may not like the notation

9e^(-0.55)(X)

It may be thinking that this is equal to

9xe^(-.55)

Try putting both in the bracket like

9e^(-.55x)

Hmm...other then that I don't know because what I wrote is correct.

10. Originally Posted by AllanCuz
Try putting both in the bracket like

9e^(-.55x)

Hmm...other then that I don't know because what I wrote is correct.
That did it. Thank you so much for all the help.

11. q''(t) = ((-272.25e^(0.55t))((e^0.55t) - 9)) / (((e^0.55t) + 9)^3) = 0

Solve for t.

Would my first step be to multiply both sides by (((e^0.55t) + 9)^3) giving me

((-272.25e^(0.55t))((e^0.55t) - 9))(((e^0.55t) + 9)^3) = 0

12. Originally Posted by WahooMan
q''(t) = ((-272.25e^(0.55t))((e^0.55t) - 9)) / (((e^0.55t) + 9)^3) = 0

Solve for t.

Would my first step be to multiply both sides by (((e^0.55t) + 9)^3) giving me

((-272.25e^(0.55t))((e^0.55t) - 9))(((e^0.55t) + 9)^3) = 0
From your prior posts it looks like you need to review multiplication my man.

Look at

((-272.25e^(0.55t))((e^0.55t) - 9)) / (((e^0.55t) + 9)^3) = 0

If you multiply the left side by the denominator it sure doesn't become

((-272.25e^(0.55t))((e^0.55t) - 9))(((e^0.55t) + 9)^3)

For example, if I have

$x/9=1$

And I multply both sides by 9, what do I get on the left side?

13. Originally Posted by AllanCuz
From your prior posts it looks like you need to review multiplication my man.

Look at

((-272.25e^(0.55t))((e^0.55t) - 9)) / (((e^0.55t) + 9)^3) = 0

If you multiply the left side by the denominator it sure doesn't become

((-272.25e^(0.55t))((e^0.55t) - 9))(((e^0.55t) + 9)^3)

For example, if I have

$x/9=1$

And I multply both sides by 9, what do I get on the left side?
9x/9 = 9 which reduces to x = 9

Wait.. so would it just be ((-272.25e^(0.55t))((e^0.55t) - 9)) = 0

14. Originally Posted by WahooMan
9x/9 = 9 which reduces to x = 9
Yes now to the same for you equations...

((-272.25e^(0.55t))((e^0.55t) - 9)) / (((e^0.55t) + 9)^3) = 0

Multiply the left by our denominator which is

(((e^0.55t) + 9)^3)

and we end up with

((-272.25e^(0.55t))((e^0.55t) - 9))

You said we ended up with

((-272.25e^(0.55t))((e^0.55t) - 9))(((e^0.55t) + 9)^3)

Notice how you multiplied the left side twice!

15. Originally Posted by AllanCuz
Yes now to the same for you equations...

((-272.25e^(0.55t))((e^0.55t) - 9)) / (((e^0.55t) + 9)^3) = 0

Multiply the left by our denominator which is

(((e^0.55t) + 9)^3)

and we end up with

((-272.25e^(0.55t))((e^0.55t) - 9))

You said we ended up with

((-272.25e^(0.55t))((e^0.55t) - 9))(((e^0.55t) + 9)^3)

Notice how you multiplied the left side twice!
Ahh I got it now.. man you are awesome.. keep an eye on this thread because I'm probably going to need some more help

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