# Math Help Equations

1. Aug 4, 2005

1) I need to express n in terms of X in this equation:

$$X = \frac {10^n-1} {10^n}$$

I got to this so far but I don't know if I'm on the right path or not and I don't know how to continue:

$$\log {X} + n = \log {(10^n-1)}$$

2) I have to prove this:

$$\frac {1} {\sqrt{4-2\sqrt {3}}} = \frac {\sqrt{3}+1} {2}$$

Don't know how to continue from here:

$$\frac {\sqrt{2} \sqrt {2+\sqrt {3}}} {2}$$

2. Aug 4, 2005

### zwtipp05

For number one:
$$X = \frac {10^n-1} {10^n}$$
$$X = 1 - \frac {1} {10^n}$$
Which gives you:
$$1 - X = \frac {1} {10^n}$$

Can you go from here?

3. Aug 4, 2005

### saltydog

For (2), note that:

$$1+\frac{\sqrt{3}}{2}=(a+b)^2$$

a and b has to be the expression you're trying to show right?

4. Aug 4, 2005

### EnumaElish

$$2 = (\sqrt{3}+1) {\sqrt{4-2\sqrt {3}}}$$
$$4 = (\sqrt{3}+1)^2 (4-2\sqrt {3})$$

Now expand the square, then expand everything, then simplify to show left = right.

5. Aug 4, 2005

This is the answer for the first one, isn't it?

$$n = -\log {(1-X)}$$

It was easy, but I don't know why I couldn't do it. Thanks a lot.

I'm having serious trouble with the second one

I suppose you got those equalities by rearranging the terms in the first equation. The thing is that the problem just says simplify:

$$\frac {1} {\sqrt{4-2\sqrt {3}}}$$

I'm sorry, it was my mistake. The second part is the answer given by the teacher.

6. Aug 4, 2005

Also, do you mean I have to replace those equations in

$$\sqrt{4+2\sqrt {3}$$

to get

$$\sqrt {(\sqrt{3}+1)^2 (4-2\sqrt {3}) + [(\sqrt{3}+1) {\sqrt{4-2\sqrt {3}}}] \sqrt {3}}$$

7. Aug 4, 2005

### VietDao29

Okay. Here is what you should do :
You know that:
$$(\alpha + \beta) ^ 2 = \alpha ^ 2 + 2\alpha \beta + \beta ^ 2$$
You will try to arrange $$\aqrt{4 - 2 \sqrt{3}}$$ into something like: $$(\alpha + \beta) ^ 2$$ Then you can easily take the square root of it.
So you have $$\alpha ^ 2 + \beta ^ 2 = 4 \mbox{, and } \alpha \beta = -\sqrt{3}$$
Can you solve for $\alpha$, and $\beta$?
Then can you solve : $$\sqrt{(\alpha + \beta) ^ 2}$$? Just remember that:
$$\sqrt{A ^ 2} = |A|$$
Viet Dao,

Last edited: Aug 4, 2005
8. Aug 5, 2005