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.
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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. $\displaystyle \left(\sqrt{x+2}\right)^2 > 0^2$ $\displaystyle x+2 > 0$ $\displaystyle x > -2$
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 $\displaystyle 0<x-2<1.$
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. Thank you.
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