How solve

[yingwuluo]

1-x*2.72^(-50*y)=100/z
1-x*2.72^(-100*y)=150/z
1-x*2.72^(-150*y)=250/z

[ettlz]

With a pen and paper? \(x=-\frac12\), \(y=-\frac{\log_{2.72}2}{50}\), \(z=50\)

[Helge Gabert]

Starting values? Any ideas what the feasible solution space should look like?

Using fsolve(), with blind values x=-.1, y=-.1,z=.1*i, I get [-.5, -1.38 . . .E-2, 50].

[AlexFekken]

(08-08-2017 03:05 PM)ettlz Wrote:  With a pen and paper? \(x=-\frac12\), \(y=-\frac{\log_{2.72}2}{50}\), \(z=50\)

Yes!

Really sad that somebody would even try a calculator or computer for a problem like this.

[yingwuluo]

solve([1-x*e^(?50*y)=100/z,1-x*e^(?100*y)=200/z,1-x*e^(?150*y)=250/z],[x,y,z])
[/b]
["[y,e^(-50*y),e^(-100*y),e^(-150*y)] is not rational w.r.t. y Erreur: Valeur Argument Incorrecte"]

fsolve([(1-x*e^(-50*y)) = (100/z),(1-x*e^(-100*y)) = (200/z),(1-x*e^(-150*y)) = (250/z)],[x,y,z],[-0.1,-0.1,-0.1*i])

undef

Above output is the result from HP Prime Graphing Calculator, something wrong with it, please help me check if there is any input mistake.

[parisse]

You can help solve by rewriting the system as a rational system

Code:

solve([(1-x*t) = (100/z),(1-x*t^2) = (150/z),(1-x*t^3) = (250/z)],[x,t,z])

then

Code:

solve(e^(-50*y)=t,y)

fsolve also works with a real starting point

Code:

fsolve([(1-x*e^(-50*y)) = (100/z),(1-x*e^(-100*y)) = (150/z),(1-x*e^(-150*y)) = (250/z)],[x,y,z],[-0.1,-0.1,0.1])

[yingwuluo]

i tried to input i in solve symbolic view，it shows that irror：invalid input，please help me fix this problem.

[Helge Gabert]

You didn't use the starting values I mentioned in my above post.

[-.1,-.1,.1*i]

Works for me.

[yingwuluo]

What are the theory for using this starting value.
i tried to input i in solve symbolic view，it shows that irror：invalid input，please help me fix this problem.

[Arno K]

I, as well as the others, I think, have not tried it in the solver-app. Switch to the command-line, I tried it in CAS, and the above mentioned entry beginning with fsolve(...) will do the job.
Arno

[yingwuluo]

Thank you all. The members here are very active in solving the problem.