1. ## Inequations

This is my first post in this forum, and I sincerely hope to be doing the right thing, posting in the right section. Here it goes.

sqrt (x+2) > 0

and

(3/2) log (x-2)<0

Thank you.

2. Originally Posted by Alvy
This is my first post in this forum, and I sincerely hope to be doing the right thing, posting in the right section. Here it goes.

sqrt (x+2) > 0

and

(3/2) log (x-2)<0

Thank you.
$\displaystyle \left(\sqrt{x+2}\right)^2 > 0^2$

$\displaystyle x+2 > 0$

$\displaystyle x > -2$

3. Originally Posted by e^(i*pi)
$\displaystyle \left(\sqrt{x+2}\right)^2 > 0^2$

$\displaystyle x+2 > 0$

$\displaystyle x > -2$
actually, $\displaystyle \sqrt{x+2}>0$ is true as long as the radicand is well defined, so it's just $\displaystyle x>-2$ and there's no need to square.

Originally Posted by Alvy
(3/2) log (x-2)<0
that's just $\displaystyle \log(x-2)<0.$

we require that $\displaystyle 0<x-2<1.$

4. Originally Posted by Krizalid
that's just $\displaystyle \log(x-2)<0.$

we require that $\displaystyle 0<x-2<1.$
I thought it was like this.

$\frac{3}{2}\log_a{(x\,-\,2)}\,\lt\,0,\,a\,\gt\,0,\,x\,\gt\,2$

$\log_a{(x\,-\,2)^{3/2}}\,\lt\,0$

$a^{\log_a{(x\,-\,2)^{3/2}}}\,\lt\,a^0$

$(x\,-\,2)^{3/2}\,\lt\,1$

$x\,-\,2\,\lt\,1$

$2\,\lt\,x\,\lt\,3$

Thank you.