# Equation cant solve

1. Jul 11, 2008

### devanlevin

how do i solve this equatio?
[(√2) +1]²¹=[(√2)-1]*[3+2(√2)]˟ˉ¹

(√2) is sqare root of 2

hhow do i find x?

how can i divide [(√2) +1]²¹ by =[(√2)-1]

2. Jul 11, 2008

### Staff: Mentor

I don't see x in your equation.

3. Jul 11, 2008

### BryanP

4. Jul 11, 2008

### Staff: Mentor

Perhaps I am missing something, or it is just my computer glitch, but I have no idea where and what the unknown is and I can't see it with this formatting neither in Opera nor in IE. I see either some square or nothing on the right - but even after checking its code (#735) I am not wiser, as by name that's "swedish grave accent"...

Is it shaped as x? Is this the equation?

$$(\sqrt 2 + 1) ^{21} = \frac {(\sqrt 2 - 1)(3+2 \sqrt 2)} {x}$$

5. Jul 11, 2008

### BryanP

I'm confused as well. It appears that the x is located as an exponent of [3 + 2*sqrt(2)], more like:

[3 + 2*sqrt(2)]^(x-1)

Last edited: Jul 11, 2008
6. Jul 11, 2008

### Staff: Mentor

Something like

$$(\sqrt 2 + 1) ^{21} = (\sqrt 2 - 1)(3+2 \sqrt 2) ^{x-1}$$

7. Jul 11, 2008

### BryanP

Yeah like that.

Sorry about what I posted earlier, I changed it since I accidently put -x instead of x-1

8. Jul 11, 2008

### devanlevin

exactly that, x-1 is the exponent

9. Jul 11, 2008

### HallsofIvy

Staff Emeritus
So your problem is simply to solve a= bcx-1 where a, b, c are numbers?

Surely you see that cx-1= (a/b). Now take the logarithm of both sides.

10. Jul 12, 2008

### gel

First , you can write
$$\frac{1}{\sqrt{2}-1}=\frac{\sqrt{2}+1}{(\sqrt{2}+1)(\sqrt{2}-1)}=\sqrt{2}+1.$$
Also,
$$(\sqrt{2}+1)^2=2+2\sqrt{2}+1=3+2\sqrt{2}.$$
Hopefully, that helps

11. Jul 12, 2008

### arildno

Your equation is of the form:
$$a=b*c^{x-1}$$
where a,b,c are ugly numbers.
Solve for x in this GENERAL form first, and then, if you absolutely have to, substitute them with your ugly numbers.

12. Jul 12, 2008

### gel

I disagree totally with this. The general answer would be x=log(a/b)/log(c)+1, but the point of the original question seems to be to do some simple arithmetic in [tex]Z[\sqrt{2}][/itex], and the answer is an integer, and can be solved exactly without resorting to log tables or a calculator. Expressed as logs, that would not be clear.