
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
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: $\left\log_xa\right=\left\{\begin{aligned}\log_x &a\left\{\begin{aligned}x>1\ &\land\ a\geq 1\\0<x<1\ &\land\ 0<a\leq 1\end{aligned}\right.\\\log_x&a\left\{\begin{aligned}x>1\ &\land\ 0<a<1\\0<x<1\ &\land\ a>1\end{aligned}\right.\end{aligned}\right.$ 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 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Logarithm inequality  Ionika  Algebra  3  February 23rd, 2015 12:07 PM 
logarithm inequality  bongantedd  Algebra  6  May 1st, 2014 06:56 AM 
logarithm inequality  bongantedd  Algebra  3  May 1st, 2014 01:18 AM 
Logarithm Inequality  sachinrajsharma  Algebra  1  February 13th, 2013 10:01 AM 
Logarithm property involving imaginary exponent  mjohnson91  Complex Analysis  1  January 27th, 2010 01:16 PM 