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May 21st, 2018, 11:59 PM  #1 
Newbie Joined: Apr 2018 From: Banovo Brdo Posts: 12 Thanks: 0  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. 

Tags 
abs, exponent, inequailty, inequality, logarith, logarithm 
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