My Math Forum Exponential function inequality
 User Name Remember Me? Password

 Algebra Pre-Algebra and Basic Algebra Math Forum

 April 29th, 2014, 04:11 PM #1 Newbie     Joined: Apr 2014 From: AQ Posts: 3 Thanks: 0 Math Focus: casual math user Exponential function inequality Hi! This is something that came up in a practical problem. 9 - 2^2x >= 0 This was my attempt at solving it with my rusty skillz: 9 - 2^2x >= 0 -2^2x >= -9 2^2x =< 9 (multiplying by -1, swapping) 2^x =< sqrt(9) (hope this is valid) 2^x =< 3 So far so good (I think...?). Now for the last step, isolating x by turning it into a log function. Considering: a^y = x equivalent to loga(x) = y we get something like 2^x =< 3 equivalent to log2(3) =< x, but that doesn't seem to make sense. I should swap the inequality again... (log2(3) >= x, or ~1,585 >= x, which seems to be the answer I want) Why do I have to swap the inequality again? Also, did I do anything wrong in the process?
 April 29th, 2014, 04:33 PM #2 Senior Member     Joined: Apr 2014 From: Europa Posts: 575 Thanks: 176 $\displaystyle 2^x\leq3 \Leftrightarrow \ \log_2 2^x\leq\log_2 3\Leftrightarrow x\leq \log_2 3.$ Thanks from Fate
 April 29th, 2014, 04:58 PM #3 Newbie     Joined: Apr 2014 From: AQ Posts: 3 Thanks: 0 Math Focus: casual math user Right, that actually makes sense with the breaking down. Thanks!

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post annakar Algebra 2 December 9th, 2012 01:31 AM arnold Calculus 4 September 22nd, 2011 12:28 PM kingcoke Calculus 2 October 8th, 2010 08:09 AM sphinn Algebra 1 September 23rd, 2010 11:46 PM arnold Real Analysis 1 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top