Solve for x

1. Feb 26, 2007

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 $a \in \mathbb{R}$ .

Last edited: Feb 26, 2007
2. Feb 26, 2007

whatta

I just did quick graph of a(x), and it looks like no real roots for 0<a<1 exist.

3. Feb 26, 2007

dextercioby

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.

4. Feb 26, 2007

murshid_islam

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

5. Feb 26, 2007

dextercioby

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

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

Solve it...

6. Feb 26, 2007

murshid_islam

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

Last edited: Feb 26, 2007
7. Feb 26, 2007

dextercioby

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. Feb 26, 2007

HallsofIvy

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. Feb 26, 2007

eaboujaoudeh

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

10. Feb 27, 2007

Gib Z

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. Feb 27, 2007

murshid_islam

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

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

Last edited: Feb 27, 2007
12. Feb 27, 2007

dextercioby

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

13. Feb 28, 2007

uart

It cant be written in terms of elementry function but it can be written in terms of the lambert W function.

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

Last edited: Feb 28, 2007
14. Feb 28, 2007

HallsofIvy

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. Mar 1, 2007

murshid_islam

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

Last edited: Mar 1, 2007
16. Mar 1, 2007

tehno

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

17. Mar 2, 2007

Gib Z

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.