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Math Help - Natural logarithm problem involving e, solve for k

  1. #1
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    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?
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by WahooMan View Post
    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
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  3. #3
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    Got it. k=0.55 Thank you so much.
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  4. #4
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    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.
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  5. #5
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by WahooMan View Post
    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.
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  6. #6
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    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?
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  7. #7
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by WahooMan View Post
    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.
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  8. #8
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    Quote Originally Posted by AllanCuz View Post
    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?
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  9. #9
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by WahooMan View Post
    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.
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  10. #10
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    Quote Originally Posted by AllanCuz View Post
    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.
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    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
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  12. #12
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by WahooMan View Post
    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?
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    Quote Originally Posted by AllanCuz View Post
    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
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  14. #14
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by WahooMan View Post
    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!
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  15. #15
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    Quote Originally Posted by AllanCuz View Post
    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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