Algebra Pre-Algebra and Basic Algebra Math Forum

 October 22nd, 2016, 02:32 AM #1 Senior Member   Joined: Jul 2011 Posts: 405 Thanks: 16 exponential inequality values of parameter $a$ for which the inequality $a\cdot9^x+4(a-1)3^x+a>1$ is true October 22nd, 2016, 05:45 AM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 This is sort of a trick question. $9^x > 0 < 4 * 3^x\ for\ all\ real\ x.$ Can you take it from here? Thanks from panky October 22nd, 2016, 06:10 AM #3 Senior Member   Joined: Jul 2011 Posts: 405 Thanks: 16 To jeffm would you like to explain me in detail, Thanks November 8th, 2016, 03:22 AM #4 Senior Member   Joined: Jul 2011 Posts: 405 Thanks: 16 Thanks Jeffm1 got it. Denoting $3^x=t>0$, we need $a>\dfrac {4t+1}{t^2 + 4t+1}$ for all $t>0$. Since the function $t\mapsto \dfrac {4t+1}{t^2 + 4t+1}$ is decreasing on $[0,\infty)$, we need $a\geq \lim_{t\to 0} \dfrac {4t+1}{t^2 + 4t+1} = 1$, and so indeed the answer is $a\in [1,\infty)$ Tags expnential, exponential, inequality 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 panky Algebra 2 June 4th, 2016 03:25 AM mared Algebra 4 September 5th, 2014 03:13 PM Fate Algebra 2 April 29th, 2014 04:58 PM annakar Algebra 2 December 9th, 2012 01:31 AM sphinn Algebra 1 September 23rd, 2010 11:46 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      