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?

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- Apr 24th 2011, 08:06 AMdumluckSimplification 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? - Apr 24th 2011, 08:26 AMProve It
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. - Apr 24th 2011, 08:30 AMSoroban
Hello, dumluck!

Quote:

. . . . . . . . . . . . . . . . . _____

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

- Apr 24th 2011, 08:52 AMProve It
- Apr 25th 2011, 05:31 AMdumluck
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.

? - Apr 25th 2011, 06:25 AMtopsquark
- Apr 25th 2011, 07:06 AMdumluck