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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. #11
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    Quote Originally Posted by nsganon101 View Post
    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
    Just for the sake of giving you some feedback you can't get when you teach yourself, we would say you "raised e to the 0 power", or something like that, not "rose up"; and you didn't make e = 1 (it's still 2.71828!). It can be important to say things clearly to avoid confusing people.

    Quote Originally Posted by nsganon101 View Post
    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
    Can you tell us what the answers are, in case their form might give us a clue what they are doing? If I were helping you in person, I would long ago have grabbed the book to see just what it says, looking for clues to point out to you. Maybe there's something you could quote for us.

    Quote Originally Posted by nsganon101 View Post
    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
    I'd say this is very much a "problem", and in fact essentially what I proposed! It's much more that just a bare equation with no context, as you presented it to us.

    As has been said, this will probably require some numerical method; if you have a graphing calculator, or can use an online equivalent, you could use it to solve the equation stapel showed (and solved with Wolfram Alpha). Unless the book gives any extra clues, that is probably the most reasonable thing to do. One of the little secrets they don't teach you in algebra class is that most equations can't be solved exactly; they usually show you only the problems that algebra can solve.

  2. #12
    Quote Originally Posted by Dr.Peterson View Post
    Just for the sake of giving you some feedback you can't get when you teach yourself, we would say you "raised e to the 0 power", or something like that, not "rose up"; and you didn't make e = 1 (it's still 2.71828!). It can be important to say things clearly to avoid confusing people.



    Can you tell us what the answers are, in case their form might give us a clue what they are doing? If I were helping you in person, I would long ago have grabbed the book to see just what it says, looking for clues to point out to you. Maybe there's something you could quote for us.



    I'd say this is very much a "problem", and in fact essentially what I proposed! It's much more that just a bare equation with no context, as you presented it to us.

    As has been said, this will probably require some numerical method; if you have a graphing calculator, or can use an online equivalent, you could use it to solve the equation stapel showed (and solved with Wolfram Alpha). Unless the book gives any extra clues, that is probably the most reasonable thing to do. One of the little secrets they don't teach you in algebra class is that most equations can't be solved exactly; they usually show you only the problems that algebra can solve.
    I gave you what I had. Also, I am trying to find approximates. Sorry I didn't clarify that first.

    The answers they gave me were k1 is roughly equal to: 2.10999, k2 is roughly equal to: 3.87937 and k3 is roughly equal to: -2.10999
    Last edited by nsganon101; 12-15-2017 at 05:54 PM.

  3. #13
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    Quote Originally Posted by nsganon101 View Post
    I gave you what I had. Also, I am trying to find approximates. Sorry I didn't clarify that first.

    The answers they gave me were k1 is roughly equal to: 2.10999, k2 is roughly equal to: 3.87937 and k3 is roughly equal to: -2.10999
    Using those numbers, I was able to find the book online, and I see why you mentioned finding the solution without using Mathematica: that's how they say they found it! And they emphasized that the answers are approximate (though their explanation is not entirely accurate). So your initial question really should have been "What is a free alternative to Mathematica?"!

    There are many tools you can use instead of Mathematica, such as a graphing calculator. You could either give the entire original system, or a simplified form from a later step. Using Wolfram Alpha, which is closely related to Mathematica, I got this. You could also use a method such as Newton's method to solve manually, but why bother when the software is available?

    You see now, I imagine, why when you ask a question it's best to provide all the information available, starting with an exact statement of the problem. (See the Read before Posting message at the top of this forum.) It can save a lot of effort.

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