Solve for x

  • #1
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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Answers and Replies

  • #2
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I just did quick graph of a(x), and it looks like no real roots for 0<a<1 exist.
 
  • #3
dextercioby
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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
dextercioby
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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
dextercioby
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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
HallsofIvy
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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
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well at least there's a limit for x....as a-->infinity....x-->1 :D
 
  • #10
Gib Z
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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
dextercioby
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Nope, it can't be done algebraically. Graph intersection.
 
  • #13
uart
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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 cant 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
HallsofIvy
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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
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
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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
Gib Z
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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.
 

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