Is x = e^{W(1)} a Solution for Any Polynomial with Rational Coefficients?

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In summary, the conversation is about solving the equation x^a - x = 1 for any value of a. By taking the natural logarithm of both sides, the solution x = 5 is found. However, it is noted that for some values of a, such as a = 4, the equation cannot be solved algebraically and numerical methods must be used. It is also mentioned that there exist polynomial equations with no solutions in terms of radicals. Finally, the conversation shifts to the equation x^x = e, which is a transcendental equation and cannot be solved algebraically.
  • #1
murshid_islam
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i have to solve for x:
[tex]x^a - x = 1[/tex] where [tex]a = \frac{\ln 6}{\ln 5}[/tex]

taking [itex]\ln[/itex] in both side, i get,
[tex]a\ln x = \ln(x+1)[/tex]

[tex]\Rightarrow\frac{\ln(x+1)}{\ln x} = a = \frac{\ln 6}{\ln 5}[/tex]

here we can see that x = 5.

but what i wanted to know is the general solution of the equation for any [itex]a \in \mathbb{R}[/itex] .
 
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  • #2
I just did quick http://www.imagehosting.com/show.php/267932_tmp.JPG, and it looks like no real roots for 0<a<1 exist.
 
  • #3
There's a simple argument for the uniqueness of the solution to the problem.

Consider the function

[tex] f(x)=\ln 5 \ln (x+1) -\ln 6 \ln x [/tex]

x=5 is clearly a solution. Since the derivative of the function is monotonic negative on [itex] \mathbb{R}_{+} [/itex], it follows that the function is everly decreasing on the positive semiaxis, therefore the x=5 solution (zero of the function) is unique.
 
  • #4
but what is the general solution for any [itex]a \in \mathbb{R}[/itex]?
 
  • #5
In the general case, the problem's more complicated, as, for example the case a=4 shows.

[tex] x+1=x^{4} [/tex]

Solve it...
 
  • #6
dextercioby said:
[tex]x+1=x^{4}[/tex]

Solve it...
its approximately x = 0.72449
i solved it by numerical methods. but how can i solve it algebraically?
 
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  • #7
The point i was trying to make is that you generally can't (solve it algebraically). Your problem was simple, but a general one isn't...
 
  • #8
You do understand, don't you, that there exist polynomial equations (or degree 5 or higher) that have NO solutions in terms of radicals?
 
  • #9
well at least there's a limit for x...as a-->infinity...x-->1 :D
 
  • #10
There exist methods of solving polynomials in terms of radicals and co-effecients, but only if the degree is 5 or lower, As Halls said. But even those methods, to me at least, are very long and cumbersome. In fact i remember seeing the equations which give x for a general quartic equation. There were 4 different equations for the 4 values of x, and each equation went at least across your entire screen, taking 2 lines.
 
  • #11
can we solve the following algebraically:
[tex]x^x = e[/tex]

i got this:
[tex]\ln(x^x) = 1[/tex]
[tex]\Rightarrow x\ln x = 1[/tex]
 
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  • #12
Nope, it can't be done algebraically. Graph intersection.
 
  • #13
murshid_islam said:
can we solve the following algebraically:
[tex]x^x = e[/tex]

i got this:
[tex]\ln(x^x) = 1[/tex]
[tex]\Rightarrow x\ln x = 1[/tex]

It can't be written in terms of elementry function but it can be written in terms of the lambert W function.

[tex] x = e^{W(1)} [/tex]
 
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  • #14
Gib Z said:
There exist methods of solving polynomials in terms of radicals and co-effecients, but only if the degree is 5 or lower, As Halls said. But even those methods, to me at least, are very long and cumbersome. In fact i remember seeing the equations which give x for a general quartic equation. There were 4 different equations for the 4 values of x, and each equation went at least across your entire screen, taking 2 lines.
My understanding was "only if the degree is lower than 5"- there exist polynomials of degree 5 which cannot be solved "by radicals" (because S5[/sup] is not a solvable group).
 
  • #15
dextercioby said:
Nope, it can't be done algebraically. Graph intersection.
how can we prove that we can't solve [tex]x^x = e[/tex] algebraically? can anyone please help?
 
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  • #16
murshid_islam said:
how can we prove that we can't solve [tex]x^x = e[/tex] algebraically? can anyone please help?


That's a transcendental equation.:smile:
Not of polynomial type.
 
  • #17
Halls: My bad sorry, you're correct of course. Abel-Ruffeni's Theorem, i think it is, just incase anyone is interested.

murshid_islam: Prove [itex] x = e^{W(1)} [/itex] isn't the solution to any polynomial with rational co-effecients.
 

What is the meaning of "x^a" in the equation?

The notation "x^a" means x raised to the power of a. In other words, x^a is the same as x multiplied by itself a times.

What is the purpose of solving x^a - x = 1 for any a?

The purpose is to find the value of x that satisfies the equation for any given exponent a. This can help in understanding the behavior of exponential functions and their relationship with linear functions.

What is the general method for solving this equation?

The general method for solving x^a - x = 1 for any a is to first isolate the x term on one side of the equation and then use logarithms to solve for x.

Can this equation be solved for irrational or complex values of a?

Yes, this equation can be solved for irrational and complex values of a. The solution may involve using complex numbers and the natural logarithm function.

What are some real-life applications of solving x^a - x = 1 for any a?

This equation has various applications in fields such as physics, chemistry, and engineering. It can be used to model population growth, decay of radioactive materials, and electrical circuits.

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