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 May 21st, 2018, 10:59 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\\01\ &\land\ 01\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 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Ionika Algebra 3 February 23rd, 2015 12:07 PM bongantedd Algebra 6 May 1st, 2014 06:56 AM bongantedd Algebra 3 May 1st, 2014 01:18 AM sachinrajsharma Algebra 1 February 13th, 2013 10:01 AM mjohnson91 Complex Analysis 1 January 27th, 2010 01:16 PM

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