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?
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.
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
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.
?