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

nsganon101

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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
 
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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.
 
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.)

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 "\(\displaystyle e^{(k_2)^0}\)", which equals "\(\displaystyle e^1\, =\, e\), so the first equation becomes:

. . . . .\(\displaystyle e\, k_1\, +\, k_3\, =\, 0\)

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

The second equation can be restated as:

. . . . .\(\displaystyle k_1\, e^{\sqrt[10]{k_2\,}}\, +\, k_3\, =\, 1\)

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

Please be complete. Thank you! ;)
 
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

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
 
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
 
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 \(\displaystyle A = k_1e^{k_2t} + k_3\), and you were given pairs (t, A) of (0, 0), (0.1, 1), and (1, 100). For clarity, I might write your equations as
\(\displaystyle k_1e^{0k_2} + k_3=0\)
\(\displaystyle k_1e^{0.1k_2} + k_3=1\)
\(\displaystyle k_1e^{k_2} + k_3=100\)

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

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.
 
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
 
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?

. . . . .\(\displaystyle 0\, =\, k_1\, e^0\, +\, k_3\)

. . . . .\(\displaystyle 0.1\, =\, k_1\, e^{k_2/10}\, +\, k_3\)

. . . . .\(\displaystyle 100\, =\, k_1\, e^{k_2}\, +\, k_3\)

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:

. . . . .\(\displaystyle \left(\dfrac{1\, +\, k_1}{k_1}\right)^{10}\, =\, \dfrac{100\, +\, k_1}{k_1}\)

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! ;)
 
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.

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.

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.
 
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
 
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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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