# Solve for x

murshid_islam
i have to solve for x:
$$x^a - x = 1$$ where $$a = \frac{\ln 6}{\ln 5}$$

taking $\ln$ in both side, i get,
$$a\ln x = \ln(x+1)$$

$$\Rightarrow\frac{\ln(x+1)}{\ln x} = a = \frac{\ln 6}{\ln 5}$$

here we can see that x = 5.

but what i wanted to know is the general solution of the equation for any $a \in \mathbb{R}$ .

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whatta
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.

Homework Helper
There's a simple argument for the uniqueness of the solution to the problem.

Consider the function

$$f(x)=\ln 5 \ln (x+1) -\ln 6 \ln x$$

x=5 is clearly a solution. Since the derivative of the function is monotonic negative on $\mathbb{R}_{+}$, it follows that the function is everly decreasing on the positive semiaxis, therefore the x=5 solution (zero of the function) is unique.

murshid_islam
but what is the general solution for any $a \in \mathbb{R}$?

Homework Helper
In the general case, the problem's more complicated, as, for example the case a=4 shows.

$$x+1=x^{4}$$

Solve it...

murshid_islam
dextercioby said:
$$x+1=x^{4}$$

Solve it...
its approximately x = 0.72449
i solved it by numerical methods. but how can i solve it algebraically?

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Homework Helper
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...

Homework Helper
You do understand, don't you, that there exist polynomial equations (or degree 5 or higher) that have NO solutions in terms of radicals?

eaboujaoudeh
well at least there's a limit for x...as a-->infinity...x-->1 :D

Homework Helper
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.

murshid_islam
can we solve the following algebraically:
$$x^x = e$$

i got this:
$$\ln(x^x) = 1$$
$$\Rightarrow x\ln x = 1$$

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Homework Helper
Nope, it can't be done algebraically. Graph intersection.

can we solve the following algebraically:
$$x^x = e$$

i got this:
$$\ln(x^x) = 1$$
$$\Rightarrow x\ln x = 1$$

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

$$x = e^{W(1)}$$

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Homework Helper
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).

murshid_islam
Nope, it can't be done algebraically. Graph intersection.
how can we prove that we can't solve $$x^x = e$$ algebraically? can anyone please help?

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tehno
how can we prove that we can't solve $$x^x = e$$ algebraically? can anyone please help?

That's a transcendental equation.
Not of polynomial type.

Homework Helper
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 $x = e^{W(1)}$ isn't the solution to any polynomial with rational co-effecients.