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Thread: 3 equations 3 variables: k_1e^{k_2^0}+k_3=0, k_1e^{k_2^(0.1)}+k_3=1...

  1. #1

    3 equations 3 variables: k_1e^{k_2^0}+k_3=0, k_1e^{k_2^(0.1)}+k_3=1...

    I have 3 equations:

    k1ek20 + k3 = 0

    k1ek20.1 + k3 = 1

    k1ek21 + k3 = 100

    I have simplified the first equation to: k1 = -k3 and the second to:
    k1ek2 + k3 = 100 and that's about the only real progress I've made. I was told the relation between k1 and k3 being inverse of each other is the key to it but everything I've tried hasn't given the right answer
    Last edited by nsganon101; 12-15-2017 at 01:00 PM.

  2. #2
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    It appears that k3 is generally missing from your equations.

    In any case, why not substitute this into the other two equations and continue your process?

    k1 and k3 are not inverses.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  3. #3
    Does k1 = -k3 not mean they're inverses?

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    Elite Member stapel's Avatar
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    Quote Originally Posted by nsganon101 View Post
    I have 3 equations:

    k1ek20 + k3 = 0

    k1ek20.1 + k3 = 1

    k1ek21 + k3 = 100

    I have simplified the first equation to: k1 = -k3 and the second to:
    k1ek2 + k3 = 100 and that's about the only real progress I've made. I was told the relation between k1 and k3 being inverse of each other is the key to it but everything I've tried hasn't given the right answer
    What is (believed to be) "the right answer"? What have you "tried"? (Please show your steps.)

    Quote Originally Posted by nsganon101 View Post
    Does k1 = -k3 not mean they're inverses?
    If you mean "additive inverses", then yes, the above equation claims that they are. But how did you obtain this result?

    For instance, I will guess that "ek20" means "[tex]e^{(k_2)^0}[/tex]", which equals "[tex]e^1\, =\, e[/tex], so the first equation becomes:

    . . . . .[tex]e\, k_1\, +\, k_3\, =\, 0[/tex]

    But how did you "simplify" this to get "k1 = -k3"?

    The second equation can be restated as:

    . . . . .[tex]k_1\, e^{\sqrt[10]{k_2\,}}\, +\, k_3\, =\, 1[/tex]

    How did you "simplify" this to get "k1ek2 + k3 = 100"?

    Please be complete. Thank you!

  5. #5
    Quote Originally Posted by stapel View Post

    But how did you "simplify" this to get "k1 = -k3"?
    I am teaching myself all this stuff so I'll preface with that.

    I took k1ek2(0) + k3 = 0 and multiplied 0 and k2 making it 0, then rose up e to 0 making e = 1 and making the equation k1 + k3 = 0

    Quote Originally Posted by stapel View Post
    How did you "simplify" this to get "k1ek2 + k3 = 100"?

    Please be complete. Thank you!
    I took k1ek2(1) + k3 = 0 and multiplied 1 and k2 which equals k2 and simplified it to k1ek2 + k3 = 100

  6. #6
    Oh and I have the answers by the way. This is actually not required by the book in using. I just would like to know so that I may be able to find 3 variable's values when using constraints in my exponential functions, without mathmatica haha

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    Quote Originally Posted by nsganon101 View Post
    I have 3 equations:

    k1ek20 + k3 = 0

    k1ek20.1 + k3 = 1

    k1ek21 + k3 = 100

    I have simplified the first equation to: k1 = -k3 and the second to:
    k1ek2 + k3 = 100 and that's about the only real progress I've made. I was told the relation between k1 and k3 being inverse of each other is the key to it but everything I've tried hasn't given the right answer
    As I read this, the original problem (which you really should have stated for us, to give us context) probably involved an equation something like [tex]A = k_1e^{k_2t} + k_3[/tex], and you were given pairs (t, A) of (0, 0), (0.1, 1), and (1, 100). For clarity, I might write your equations as
    [tex]k_1e^{0k_2} + k_3=0[/tex]
    [tex]k_1e^{0.1k_2} + k_3=1[/tex]
    [tex]k_1e^{k_2} + k_3=100[/tex]

    You've seen that the first tells you that [tex]k_1 + k_3=0[/tex], so you can replace k3 with -k1 everywhere:
    [tex]k_1e^{0.1k_2} - k_1=1[/tex]
    [tex]k_1e^{k_2} - k_1=100[/tex]

    I might now solve one of these for k2 (in terms of k1) and plug that into the other. Or it might be helpful to eliminate k1 by dividing one by the other!

    Now, it isn't obvious that this can be solved exactly by algebra. It will be especially helpful at this point if you can tell us the exact wording of the original problem, rather than dropping us into the middle of your work with nothing to go by. You may have incorrect equations, or the instructions may call for something other than an algebraic solution, or give a hint as to a method to be used.

  8. #8
    Quote Originally Posted by Dr.Peterson View Post
    As I read this, the original problem (which you really should have stated for us, to give us context) probably involved an equation something like
    A = k1ek2t + k3
    and you were given pairs (t, A) of (0, 0), (0.1, 1), and (1, 100). For clarity, I might write your equations as

    Now, it isn't obvious that this can be solved exactly by algebra. It will be especially helpful at this point if you can tell us the exact wording of the original problem, rather than dropping us into the middle of your work with nothing to go by. You may have incorrect equations, or the instructions may call for something other than an algebraic solution, or give a hint as to a method to be used.
    It is not a problem actually it's an equation that was given to me in this textbook I'm studying. The function is g(c) = k1ek2c + k3

    The function is supposed to be used for calculating damage with respect to taking it (in a video game) and (0,0) (0.1,1) (1,100) are the constraints put on (as in, 0 damage has 0 extra hotpoints, 10% damage gets 1 extra hitpoint and 100% damage gets 100 extra hitpoints.) He never explains how to find the three variables and I would like to know how to do that

  9. #9
    Elite Member stapel's Avatar
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    Cool

    Quote Originally Posted by nsganon101 View Post
    The function is g(c) = k1ek2c + k3
    Does this mean that the "c" is multiplied on the k2, rather than being an exponent on k2? So the system is as follows?

    . . . . .[tex]0\, =\, k_1\, e^0\, +\, k_3[/tex]

    . . . . .[tex]0.1\, =\, k_1\, e^{k_2/10}\, +\, k_3[/tex]

    . . . . .[tex]100\, =\, k_1\, e^{k_2}\, +\, k_3[/tex]

    Quote Originally Posted by nsganon101 View Post
    He never explains how to find the three variables and I would like to know how to do that
    Unfortunately, it is not reasonably feasible to attempt to provide classroom instruction within this environment. To learn how to solve systems of non-linear equations, try some lessons from this listing.

    Note: This may not be solvable algebraically; any solution may require numerical methods. For instance, I ended up with the following equation:

    . . . . .[tex]\left(\dfrac{1\, +\, k_1}{k_1}\right)^{10}\, =\, \dfrac{100\, +\, k_1}{k_1}[/tex]

    It is possible to obtain an approximate solution (here), but there's no way I'm gonna expand the binomial on the left-hand side and try to obtain a better (that is, an "exact") value!

  10. #10
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    Quote Originally Posted by nsganon101 View Post
    Does k1 = -k3 not mean they're inverses?
    Okay they are "Additive Inverses", but one normally would not drop the "Additive". I'm willing to submit to other opinions.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

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