Solving 3 equations with 3 unknowns

Question

Hamza 2013-3-5

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Dear members;

I would like to solve 3 equations using matlab. following are the 3 equations.

0972=ve*ln(mo/(mo-1010mp))
11285=ve*ln((mo-1010mp)/(mo-2145mp))
10953=ve*ln((mo-2145mp)/(mo-3395mp))

The unknowns in these equations are ve, mo and mp. is there a way to solve for the unknowns?

Thanks in advance

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Answer 1

Azzi Abdelmalek 2013-3-5

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syms me ve mo mp
eq1=0.0972-ve*log(mo/me)
eq2=0.11285-ve*log((mo-1010*mp)/(mo-2145*mp))
eq3=0.10953-ve*log((mo-2145*mp)/(mo-3395*mp))
solve(eq1,eq2,eq3,ve,mo,mp)

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Hamza 2013-3-5

thanks But i guess i don't have the symbolic maths toolbox. can you post me the answers please?

And you can replace 'me' with '(mo-1010mp)' Thanks

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Answer 2

Youssef Khmou 2013-3-5

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hi try ,

 D=solve('x*log(y/a)=0.0972','x*log((y-1010*z)/(y-2145*z))=0.11285','x*log((y-2145*z)/(y-3395*z))=0.10953','x','y','z')

what is me? it is replaced with a

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Walter Roberson 2013-3-6

编辑：Walter Roberson 2013-3-6

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One solution is

mo = me, mp = 0, ve = infinity

There might be other solutions; I have my system checking.

... There do not appear to be any other real solutions.

Hamza 2013-3-6

these results are based on the new first equation which I have edited?

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Answer 3

Roger Stafford 2013-3-6

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I claim there are either no solutions or infinitely many solutions to your three equations, and I'll tell you why, Hamza. By manipulating the equations appropriately one can rewrite them in the form:

 exp(0.0972/ve)  = 1/(1-1010*r)
 exp(0.11285/ve) = (1-1010*r)/(1-2145*r)
 exp(0.10953/ve) = (1-2145*r)/(1-3395*r)

where r = mp/mo. As you can see from this, you have in effect three equations but only two unknowns, namely 've' and 'r', and they are unlikely to have a solution. You can find out if that is so by solving, say, just the first two of these equations. These two can be expressed as:

 1-1010*r = exp(-0.0972/ve)
 1-2145*r = (1-2145*r)/(1-1010*r) * (1-1010*r) 
          = exp(-0.11285/ve)*exp(-0.0972/ve)
          = exp(-.21005/ve)

Solving for r in each equation gives

 r = (1-exp(-0.0972/ve))/1010 = (1-exp(-.21005/ve))/2145
 2145*(1-exp(-0.0972/ve)) = 1010*(1-exp(-.21005/ve))
 1010*exp(-.21005/ve)-2145*exp(-0.0972/ve)+1135 = 0

You can use matlab's 'fzero' to solve this last equation for 've' and from that determine 'r'. As I say, any such solutions are unlikely to also satisfy the third equation above. If they do, then you will have infinitely many solutions corresponding to all the ways of satisfying r = mp/mo.