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

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

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].

(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.

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.

You can help solve by rewriting the system as a rational system
then fsolve also works with a real starting point

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

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

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

Works for me.

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.

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

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