112. The equivalence relation ".s.s."

by Virgil Nicula, Sep 12, 2010, 1:46 AM

Quote:
Proposed problem. Solve the equation $x^{1+\log_a x}\ge a^2 x$ , where $a\in (0,1)\cup (1,\infty )$ and $x>0$ .

Definition. Let $x,y$ be two real numbers. Then $x\ \mathrm{.s.s.}\ y\Longleftrightarrow \mathrm{sign}(x)=\mathrm{sign}(y)\ \Longleftrightarrow\ xy>0\ \vee\ x=y=0$ , i.e. the real numbers $x$ , $y$ have same sign.

Examples.

$1.1.\ x\in [a,b]\cup [b,a]\Longleftrightarrow (x-a)(x-b)\le 0\Longleftrightarrow \left| x-\frac {a+b}{2}\right|\le \left| \frac {a-b}{2}\right|\ ;$
$1.2.\ \frac xy\ \mathrm{.s.s.}\ xy\ ;\ (|a|-|b|)\ \mathrm{.s.s.}\ (a^2-b^2\ ;\ |a|\ \mathrm{.s.s.}\ a^2\ ;\ \left( \sqrt [n] a-\sqrt [n] b\right)\ \mathrm{.s.s.}$ $(a-b)\ ;\ \sqrt [n] a\ \mathrm{.s.s.}\ a\ .$

$2.1.\ (a^x-a^y)\ \mathrm{.s.s.}\ (a-1)(x-y)$ for any $\{x,y\}\subset \mathbb R\ ,\ 0<a\ne 1\ ;$
$2.2.\ (a^x-b^x)\ \mathrm{.s.s.}\ x(a-b)$ for any $x\in \mathbb R,\ 0<a\ne 1\ .$

$3.1.\ (\log_a x-\log_a y)\ \mathrm{.s.s.}\ (a-1)(x-y)$ for any $0<a\ne 1,\ x,y>0\ ;$
$3.2.\ (\log_a x-\log_b x)\ \mathrm{.s.s.}\ [(a-1)(b-1)(x-1)(b-a)]$ for any $0<a,b\ne 1\ ,\ 0<x\ ;$
$3.3.\ \log_a x\ \mathrm{.s.s.}\ (a-1)(x-1)\ .$

$4.1.\ (\arcsin x-\arcsin y)\ \mathrm{.s.s.}\ (x-y)\ ,\ \arcsin x\ \mathrm{.s.s.}\ x\ ;$
$4.2.\ (\arccos x-\arccos y)\ \mathrm{.s.s.}\ (y-x)\ ,\ \arccos x\ \mathrm{.s.s.}\ (1-x)\ ;$
$4.3.\ (\arctan x-\arctan y)\ \mathrm{.s.s.}\ (x-y),\ \arctan x\ \mathrm{.s.s.}\ x\ .$

Exercise. $0<a,b\ne 1\Longrightarrow E\equiv (a^b-a)(b^a-b)> 0$. Indeed$,\ E\ \mathrm{.s.s.}\ (a-1)(b-1)(b-1)(a-1)=(a-1)^2(b-1)^2>0\ .$

My solution for the proposed problem. We have $0<a,x\ne 1\ .$ I note $E\equiv x^{\log_a x} -a^2\ .$ Thus, $x^{\log_ax}=a^{\log^2_ax}$ and

$E=a^{\log^2_a x}-a^2\ \mathrm{.s.s.}\ (a-1)(\log_a ^2 x-2)=(a-1)(\log_a x-\sqrt 2)(\log_a x-\sqrt 2)=$

$(a-1)\left(\log_a x-\log_a a^{\sqrt 2}\right)\left(\log_a x -\log_a \frac 1{a^{\sqrt 2}}\right)$ $\mathrm{.s.s.}$ $(a-1)^3 \left(x- a^{\sqrt 2}\right)\left( x-\frac 1{a^{\sqrt 2}}\right)$ $a^{\sqrt 2}-\frac {1}{a^{\sqrt 2}}=a^{\sqrt 2}-a^{-\sqrt 2}$ $\mathrm{.s.s.}\ (a-1)\ .$

Therefore, the solution of this equation is $S=\left\{\begin{array}{ccc} (0,\infty ) & \mathrm{if} & a=1\\\\ \left [a^{\sqrt 2},\ a^{-\sqrt 2}\right] & \mathrm{if} & 0<a<1\\\\ \left(0,a^{-\sqrt 2}\right] \cup \{ 1\} \cup \left[ a^{\sqrt 2},\infty\right) & \mathrm{if} & 1<a\end{array}\right\|$ .

Home-work. Solve the inequation $\left( 2x-\sqrt {x+3}\right)\cdot \left( \sqrt [3] {x+2}+x\right)\cdot \left[ \log_{x+3} \left(2-\sqrt {x+1}\right)\right]$ $\cdot \left[\left(2-x\right)^{x-1}-x^{2(x-1)}\right]\ge 0\ .$
This post has been edited 15 times. Last edited by Virgil Nicula, Nov 26, 2015, 9:33 PM

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    by mrmath0720, Sep 28, 2020, 1:11 AM

  • wow... i am lost.

    369302 views!

    -piphi

    by piphi, Jun 10, 2020, 11:44 PM

  • That was a lot! But, really good solutions and format! Nice blog!!!! :)

    by CSPAL, May 27, 2020, 4:17 PM

  • impressive :D
    awesome. 358,000 visits?????

    by OlympusHero, May 14, 2020, 8:43 PM

72 shouts
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  • Total entries: 456
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