Equation is as follows...
1 = ( e^(2/(3+x))) + ( e^(5/(6+x)))
Conditions--
No iterations. make any suitable assumptions , expansions etc...but no iterations. The value of X exists or not does not matter. I am looking for the method to get a direct relation so that we can solve for x without iteration.
1 = ( e^(2/(3+x))) + ( e^(5/(6+x)))
Conditions--
No iterations. make any suitable assumptions , expansions etc...but no iterations. The value of X exists or not does not matter. I am looking for the method to get a direct relation so that we can solve for x without iteration.
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Let y = 5/(6+x) - 2/(3+x). Let u = 2/(3+x). This gives y = 3(x+1)/[(x+3)(x+6)] = 3/2 * u * (x+1)/(x+6). Let v = 3/2 * (x+1)/(x+6), so that y = uv and 5/(6+x) = y + u. The equation is then
1 = e^u + e^(y + u)
= e^u + e^(uv + u)
= e^u + (e^u)^(v+1)
Letting w = e^u and z = v+1, this is still
1 = w + w^z
The only time this has a remotely decent solution is when z is an integer (or *maybe* when it's rational). But z is a rational polynomial in x, so is rational only if x itself is rational. From the numeric solution, it is extremely unlikely that x is rational. So, z0 = z(x0) where x0 is the solution of the given equation is irrational. The equation
1 = w + w^z0
is then satisfied by w0 = w(x0). The solution for w in terms of z0 is then an awful special function, without a pretty form for z0 irrational. This all strongly suggests that no solution exists in terms of standard functions. Also, Wolfram Alpha isn't able to get anything from it besides a numeric solution, and it's better at solving these sort of problems exactly than anything else I've seen, including humans.
1 = e^u + e^(y + u)
= e^u + e^(uv + u)
= e^u + (e^u)^(v+1)
Letting w = e^u and z = v+1, this is still
1 = w + w^z
The only time this has a remotely decent solution is when z is an integer (or *maybe* when it's rational). But z is a rational polynomial in x, so is rational only if x itself is rational. From the numeric solution, it is extremely unlikely that x is rational. So, z0 = z(x0) where x0 is the solution of the given equation is irrational. The equation
1 = w + w^z0
is then satisfied by w0 = w(x0). The solution for w in terms of z0 is then an awful special function, without a pretty form for z0 irrational. This all strongly suggests that no solution exists in terms of standard functions. Also, Wolfram Alpha isn't able to get anything from it besides a numeric solution, and it's better at solving these sort of problems exactly than anything else I've seen, including humans.
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You cannot really solve this for x in the traditional sense, so that you have all the x on one side, outside of every natural log and exponential.
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Genie: Hehehehe.....blow....
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See:
http://www.wolframalpha.com/input/?i=1+%…
http://www.wolframalpha.com/input/?i=1+%…