Math Help: Solving Equations with Logarithms and Square Roots

[LinkMage]

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}

 

[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?

 

[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?

 

[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.

 

[LinkMage]

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

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.

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.

 

[LinkMage]

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}}

 

[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,

 

[LinkMage]

I finally did it. Thanks a lot Viet Dao. I would never have guessed I had to do that.