How to Solve the Equation [(√2) +1]²¹ = [(√2)-1]*[3+2(√2)]˟ˉ¹ for x?

[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]

 

[Borek]

I don't see x in your equation.

 

[BryanP]

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]

there's your x Borek.

 

[Borek]

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?

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

 

[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)

 

[Borek]

Something like

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

 

[BryanP]

Something like

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

Yeah like that.

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

 

[devanlevin]

exactly that, x-1 is the exponent

 

[HallsofIvy]

So your problem is simply to solve a= bc^x-1 where a, b, c are numbers?

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

 

[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

 

[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.

 

[gel]

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.

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.[/tex]