Solving 3 equations with 3 unknowns

Dear members;

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

0.0972=ve*ln(mo/(mo-1010mp))

0.11285=ve*ln((mo-1010mp)/(mo-2145mp))

0.10953=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

Answer 1

[Azzi Abdelmalek]

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)

[Hamza]

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

Answer 2

[Youssef Khmou]

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

[Hamza]

sorry I've edited my question and replaced 'me'

i'm getting the following error message

Undefined function 'solve' for input arguments of type 'char'.

[Youssef Khmou]

ok, then Symbolic Math toolbox is missing ...

[Hamza]

oh! Can you post me the results if you have it? Thanks

[Youssef Khmou]

The solution could not be found ! check again the equations .

[Walter Roberson]

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]

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

Answer 3

[Roger Stafford]

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.