1. Simplification 2

Hi All,

[tex]{3}^{x} = {3}^{-x} = \sqrt{b+1}[\MATH]

Simplifies to [tex]{3}^{2x} + 2.3^{x}.3^{-x} + 3^{-2x} = b + 2[\MATH]

but how do we get .. 2.3^{x}.3^{-x} in the above? Would it not just be {3}^{2x} + 3^{-2x} = b +2?

2. You clearly have a typo, since there are three equalities in your first line...

Assuming that's actually 3^x + 3^{-x} = \sqrt{b + 1}

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

3^{2x} + 2(3^x)(3^{-x}) + 3^{-2x} = b + 1

3^{2x} + 2(1) + 3^{-2x} = b + 1

3^{2x} + 3^{-2x} = b - 1

So the simplification you are given is also incorrect.

3. Hello, dumluck!

. . . . . . . . . . . . . . . . . _____
3^{x} + 3^{-x} .= .√b + 1

Simplifies to: .3^{2x} + 2*3^{x}*3^{-x} + 3^{-2x} .= .b + 1

but how do we get: .2*3^{x}*3^{-x} in the above?

Would it not just be: .3^{2x} + 3^{-2x} .= .b +1 ? . certainly not!

Hey, you know better! . (a + b)^2 . .a^2 + b^2

Fact: .(a + b)^2 .= .a^2 + 2ab + b^2

4. Originally Posted by Soroban
Hello, dumluck!

Hey, you know better! . (a + b)^2 . .a^2 + b^2

Fact: .(a + b)^2 .= .a^2 + 2ab + b^2

Actually (a + b)^2 = a^2 + b^2 if and only if a = 0 or b = 0...

5. Originally Posted by Soroban
Hello, dumluck!

Hey, you know better! . (a + b)^2 . .a^2 + b^2

Fact: .(a + b)^2 .= .a^2 + 2ab + b^2

Ah yes, so in the example above (3^x + 3^{-x})^2

It would be (3^x + 3^-x)(3^x + 3^-x)

So (3^x)(3^x) + (3^x)(3^-x) + (3^-x)(3^x) + (3^-x)(3^-x)

and as prove it denoted this translates to.

3^{2x} + 2(3^x)(3^{-x}) + 3^{-2x}

3^{2x} + 2(1) + 3^{-2x} = b + 1 :

Where did the two disappear too, and why has the sign of the 1 changed in the next step?

3^{2x} + 3^{-2x} = b - 1

thx.

?

6. Originally Posted by dumluck
Ah yes, so in the example above (3^x + 3^{-x})^2

It would be (3^x + 3^-x)(3^x + 3^-x)

So (3^x)(3^x) + (3^x)(3^-x) + (3^-x)(3^x) + (3^-x)(3^-x)

and as prove it denoted this translates to.

3^{2x} + 2(3^x)(3^{-x}) + 3^{-2x}

3^{2x} + 2(1) + 3^{-2x} = b + 1 :

Where did the two disappear too, and why has the sign of the 1 changed in the next step?

3^{2x} + 3^{-2x} = b - 1

thx.

?
3^{2x} + 2(1) + 3^{-2x} = b + 1

3^{2x} + 2(1) + 3^{-2x} - 2 = b + 1 - 2

3^{2x} + 3^{-2x} = b - 1

-Dan

7. Originally Posted by topsquark
3^{2x} + 2(1) + 3^{-2x} = b + 1

3^{2x} + 2(1) + 3^{-2x} - 2 = b + 1 - 2

3^{2x} + 3^{-2x} = b - 1

-Dan
of course.. thanks.