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May 21st, 2018, 10:59 PM   #1
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From: Banovo Brdo

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Inequality with an abs. value, exponent and logarithm

This given problem is $x^\left|\log_xa\right|\leq\frac{1}{a}$ with the following solution sets: $x\in(0,1)\cup(1,+\infty),\ a\in(0,1]\\
x\in(0,1),\ a\in(1,+\infty)$
This is how I've approached it:
$\left|\log_xa\right|=\left\{\begin{aligned}\log_x &a\left\{\begin{aligned}x>1\ &\land\ a\geq 1\\0<x<1\ &\land\ 0<a\leq 1\end{aligned}\right.\\-\log_x&a\left\{\begin{aligned}x>1\ &\land\ 0<a<1\\0<x<1\ &\land\ a>1\end{aligned}\right.\end{aligned}\right.$

Then I've solved for the absolute value:
$\begin{aligned}x^{\log_xa}\leq\frac{1}{a}&\qquad \lor &&(x^{\log_xa})^{-1}\leq \frac{1}{a}
\\a\leq\frac{1}{a}& &&\frac{1}{a}\leq \frac{1}{a}\end{aligned}$

And concluded $x\in(0,1),\ a\in(0,1]\ \lor\ x\in(0,1)\cup(1,+\infty),\ a\in(0,1)\cup(1,+\infty)$ which is almost the complete opposite of the true solution.

Where did I go wrong? Thanks.
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